New Jersey Center for Teaching and Learning. Progressive Mathematics Initiative

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1 Slide 1 / 240 New Jersey Center for Teaching and Learning Progressive Mathematics Initiative This material is made freely available at and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website:

2 Slide 2 / 240 Geometry Trigonometry of Right Triangles

3 Slide 3 / 240 Table of Contents Pythagorean Theorem Similarity in Right Triangles Special Right Triangles Trigonometric Ratios Solving Right Triangles Angles of Elevation and Depression Law of Sines and Law of Cosines Area of an Oblique Triangle Click on a Topic to go to that section

4 Slide 4 / 240 Pythagorean Theorem Return to the Table of Contents

5 Slide 5 / 240 Before learning about similar right triangles and trigonometry, we need to review the Pythagorean Theorem and the Pythagorean Theorem Converse.

6 Slide 6 / 240 Recall that a right triangle is a triangle with a right angle. hypotenuse leg leg The sides form that right angle are the legs. The side opposite the right angle is the hypotenuse. The hypotenuse is also the longest side.

7 Slide 7 / 240 Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. leg2 + leg2 = hypotenuse2 or a2 + b2 = c2 a c b

8 Slide 8 / 240 Example: x 9 12 Find the length of the missing side of the right triangle. Is the missing side a leg or the hypotenuse of the right triangle?

9 Slide 9 / 240 Solve for x: = x2 x = x2 225 = x2 15 = x -15 is a extraneous solution, a distance can not equal a negative number. x = 15 12

10 Slide 10 / 240 x Find the length of the missing side of the right triangle. Is the missing side a leg or the hypotenuse of the right triangle? Example:

11 Slide 11 / The missing side is the of the right triangle. B hypotenuse 6 9 x A leg

12 Slide 12 / Find the length of the missing side. 6 9 x

13 Slide 13 / 240 A leg B hypotenuse x The missing side is the of the right triangle.

14 Slide 14 / 240 x Find the length of the missing side.

15 Slide 15 / 240 Real World Application The safe distance of the base of the ladder from a wall it leans against should be one-fourth of the length of the ladder. Thus, the bottom of a 28-foot ladder should be 7 feet from the wall. How far up the wall will a ladder reach?? 28 feet 7 feet

16 Slide 16 / ? 28 feet 7 feet The ladder will reach feet up the wall safely. 2 Solve using a + b = c

17 Slide 17 / 240 Real World Application 50 x The dimensions of a high school basketball court are 84' long and 50' wide. What is the length from one corner of the court to the opposite corner? 84

18 Slide 18 / A NBA court is 50 feet wide and the length from one corner of the court to the opposite corner is feet. How long is the court? A feet B feet C 118 feet D 94 feet (Round the answer to the nearest whole number)

19 Slide 19 / 240 Pythagorean Theorem Applications The Pythagorean Theorem can also be used in figures that contain right angles.

20 Slide 20 / 240 Example Find the perimeter of the square. 18 cm Psq = 4s note: Before finding the perimeter of the square, we need to first find the length of each side.

21 Slide 21 / 240 Remember, in a square all sides are congruent. 18 cm x Start here: 2 x 2 + x 2 = 18

22 Slide 22 / 240 Example Find the area of the triangle. The base of the triangle is given, but we need to find the height of the triangle. A = 1 2 bh 13 feet 13 feet 10 feet

23 Slide 23 / feet h 5 feet 13 feet 5 feet By definition, the altitude (or height) of an isosceles triangle is the perpendicular bisector of the base.

24 Slide 24 / 240 Try this... Find the perimeter of the rectangle. 10 in P rect = 2l + 2w 8 in

25 Slide 25 / Find the area of the rectangle. B 84 square feet C 46 square inches D 46 square feet 8 feet 1 ee 7f t A 120 square feet

26 Slide 26 / Find the perimeter of the square. (Round to the nearest tenth) D 36 cm cm C 25.6 cm 9 B 25.5 cm A 12.8 cm

27 Slide 27 / inches 7 inches 10 inches 8 Find the area of the triangle.

28 Slide 28 / Find the area of the triangle. 7 inches 4 inches 7 inches

29 Slide 29 / 240 Converse of the Pythagorean Theorem If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. B If c = a + b, then ABC is a right triangle a C c b A

30 Slide 30 / 240 Example Tell whether the triangle is a right triangle. Remember c is the longest side E 7 25 F D 24

31 Slide 31 / 240 Theorem If the square of the longest side of a triangle is greater than the sum of the squares of the other two sides, then the triangle is obtuse. B If c > a + b, then ABC is obtuse. 2 2 c a 2 C b A

32 Slide 32 / 240 Theorem If the square of the longest side of a triangle is less than the sum of the squares of the other two sides, then the triangle is acute. B a c If c2 < a2 + b2, then ABC is acute. A b C

33 Slide 33 / 240 Example Classify the triangle as acute, right, or obtuse

34 Slide 34 / Classify the triangle as acute, right, obtuse, or not a triangle. 12 B right C obtuse D not a triangle A acute

35 Slide 35 / 240 A acute B right C obtuse D not a triangle Classify the triangle as acute, right, obtuse, or not a triangle.

36 Slide 36 / 240 A acute 25 B right C obtuse D not a triangle Classify the triangle as acute, right, obtuse, or not a triangle.

37 Slide 37 / Tell whether the lengths 35, 65, and 56 represent the sides of an acute, right, or obtuse triangle. B right C obtuse A acute

38 Slide 38 / 240 A acute triangle B right triangle C obtuse triangle 14 Tell whether the lengths represent the sides of an acute, right, or obtuse triangle.

39 Slide 39 / 240 Review If c = a + b, then triangle is right If c2 > a2 + b2, then triangle is obtuse. If c2 < a2 + b2, then triangle is acute.

40 Slide 40 / 240 Similarity in Right Triangles Return to the Table of Contents

41 Slide 41 / 240 There are many proofs to the Pythagorean Theorem. How many do you know? Triangle similarity can be used to prove the Pythagorean Theorem. How?

42 Slide 42 / 240 Theorem The altitude of a right triangle divides the triangle into two smaller triangles that are similar to the original triangle and each other. C A B D CD is the altitude of ABC~ ACD~ ABC CBD

43 Slide 43 / 240 Teacher Notes To prove this, click for Lab 1 - Similar Right Triangles Therefore, the altitude of a right triangle divides the triangle into two smaller triangles that are similar to the click original triangle and similar to each other. click

44 Slide 44 / 240 C Let's prove the Theorem. The altitude of a right triangle divides the triangle into two smaller triangles that are similar to the original triangle and each other. Given: Prove: ABC is a right triangle is the altitude of ABC ABC~ ACD~ CBD A B D Statements ABC is a right triangle is a right angle Reasons Given click Given click Def clickof Altitude is a right angle Def of Perp Lines. 2 lines that form a rtclick angle All rt angles are click Reflexive Prop of click ABC ~ ACD is a right angle AA~ click Def of Perp Lines click All rt angles are click Reflexive Prop of click ABC ~ CBD ABC~ ACD~ CBD AA~ click Transitive Prop of ~ click

45 Slide 45 / 240 C Let's sketch the 3 triangle's separately, with the same orientation. A B B C C A A D C B D Match up the angles. D Helpful tip: If you set, then you can assign all the angles a value and easily find the matches. 30 B B C A C A 30 D C D

46 Slide 46 / 240 e c a Assign lengths to all the segments. Let the lengths of the segments on the hypotenuse be d and e. B B c d b Label the sides of a triangle with the lower case letter of the opposite angle. A b C a C a b A d D C e ABC~ D ACD~ Because the triangles are similar the corresponding sides are proportional. ABC~ ACD ABC~ CBD CBD

47 Slide 47 / 240 To prove the Pythagorean Theorem, use the proportions. Given: ABC~ ACD click Using the multiplication property of equality, multiply the equation by bc. click Prove: (1) e a Altitude of a rt triangle click theorem. Definition of similar triangles. ABC is a right triangle. is an altitude. c Reasons Statements ABC~ CBD simplify click Altitude of a rt triangle click theorem. Definition of similar triangles. click d Using the multiplication property of equality, click multiply the equation by ac b (2) simplify click

48 Slide 48 / 240 To prove the Pythagorean Theorem, use the proportions (continued). Given: ABC is a right triangle. is an altitude. Prove: Statements Reasons Using the addition property of equality, add equation (1) and equation (2) together. click Distributive Property click click Given Substitution click e c d b a Simplify click

49 Slide 49 / 240 Example Find the length of the altitude KI? H K I 5 J

50 Slide 50 / 240 H It maybe helpful to sketch the 3 triangle's separately, with the same orientation. H H I x K K 5 5 J J 12 x I K x 13 I J 12 5 K Because the triangles are similar the corresponding sides are proportional. 13x = 60 x 4.62

51 Slide 51 / 240 Try this... Find the length of RS. 3 P Q R 4 5 x S 5 P x 4 3 R R Q S 4 Q P 5 S S

52 Slide 52 / Which ratio is the ratio of corresponding sides? A B C D H K J I

53 Slide 53 / Find KJ. I 7 J K 25 H 24

54 Slide 54 / 240 The next two theorems are Geometric Mean Theorems. What is a mean? An average. Usually when we ask to find the mean, we are asking for the arithmetic mean. What is an arithmetic mean? The sum of n values divided by the number of values (n). What is a geometric mean? The nth root of a product of n values. It is defined for only positive numbers (no negative numbers, no zero) For more information click on this link: Arithmetic Mean vs Geometric Mean

55 Slide 55 / 240 The geometric mean of two positive numbers a and b is the positive number x that satisfies a =x x b x2 = ab x= Visually, the geometric mean answers this question: given a rectangle with sides a and b, find the side of the square whose area equals that of the rectangle.

56 Slide 56 / 240 Example Find the geometric mean of 8 and 14. x2 = 8(14) x2 = 112 (only the positive value)

57 Slide 57 / Find the geometric mean of 7 and 56. Write the answer is simplest radical form. B C D A

58 Slide 58 / Find the geometric mean of 3 and 48. Students type their answers here

59 Slide 59 / 240 Corollary The altitude drawn to the hypotenuse of a right triangle divides the the hypotenuse into two segments. The altitude is the geometric mean of the two segments formed. C CD is the altitude of Since, A D ACD~ CBD B CD2 = AD(DB) ABC

60 Slide 60 / 240 Example Find z. 8 6 z

61 Slide 61 / 240 Example Find z z

62 Slide 62 / 240 Try this... Find y y 2) ) y

63 Slide 63 / Find x. 10 C 20 D 50 5 B 100 x A

64 Slide 64 / Find x. A 99 C D 9 x B 11

65 Slide 65 / 240 Corollary If the altitude drawn to the hypotenuse of a right triangle, divides the hypotenuse into two segments. The length of each leg of the original triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. C A D CD is the altitude of B Since, ABC~ ABC~ ACD AD AC = AC AB ABC ACD~ CBD ABC~ CBD AB BC = BC DB

66 Slide 66 / 240 Example 4 R S 9 T Find x. x U

67 Slide 67 / 240 Example D E 4 F 6 G Find x. x

68 Slide 68 / Is PR a geometric mean between QR and SR? True False Q S R P

69 Slide 69 / Is the geometric mean correct? P False Q S R True

70 Slide 70 / Which proportion is correct? K L M A B C D J

71 Slide 71 / Find y. B 20 C 5 D 12 y 9 16 A

72 Slide 72 / Find y. B 18 C 24 D None of the above 27 y x 9 A 3

73 Slide 73 / Find x. 5 8 x

74 Slide 74 / 240 Special Right Triangles Return to the Table of Contents

75 Slide 75 / 240 In this section you will learn about the properties of the two special right triangles o 90o 60o 45o 90o 30o

76 Slide 76 / Triangle Theorem hypotenuse = leg( 2) 45o Can you prove this? x 2 x x 45o A triangle is an isosceles right triangle, where the hypotenuse is 2 times the length of the leg.

77 Slide 77 / 240 P 6 Q 45o y x 45o R Example Find the length of the missing sides. Write the answer in simplest radical form.

78 Slide 78 / 240 y S T x 18 V Example Find the length of the missing sides of the right triangle.

79 Slide 79 / 240 Try this... Find the length of the missing sides. y 8 x

80 Slide 80 / Find the value of x. A 5 C (5 2)/2 y 5 B 5 2 x

81 Slide 81 / Find the value of y. A 5 B 5 2 C (5 2)/2 y 5 x

82 Slide 82 / What is the length of the hypotenuse of an isosceles right triangle, if the length of the legs is 8 2 inches.

83 Slide 83 / What is the length of each leg of an isosceles, if the length of the hypotenuse is 20 cm.

84 Slide 84 / Triangle Theorem In a right triangle, the hypotenuse is twice the length of the shorter leg and the longer leg is 3 times the length of the shorter leg. x 60o hypotenuse = 2(shorter leg) longer leg = 3(shorter leg) 2x 30o x 3

85 Slide 85 / 240 This can be proved using an equilateral triangle. x For right triangle ABD, BD is a perpendicular bisector. let a = x, c = 2x and b= BD B c=2x A 60 a=x 2x x x b D 60o x 60 C 30o

86 Slide 86 / 240 G 30o Example Find the length of the missing sides of the right triangle. x y H 60o 5 F

87 Slide 87 / 240 G x HF is the shortest side GF is the longest side (hypotenuse) GH is the 2nd longest side HF < GH < GF 30o Recall triangle inequality, the shortest side is opposite the smallest angle and the longest side is opposite the largest angle. y H 60o 5 F

88 Slide 88 / 240 x M 60o y 9 30o T A Example Find the length of the missing sides of the right triangle.

89 Slide 89 / 240 Example Find the area of the triangle. 14 ft

90 Slide 90 / 240 The altitude (or height) divides the triangle into two 30o-60o-90o triangles. h? The length of the shorter leg is 7 ft. The length of the longer leg is 7 3 ft. A = b(h) = 14(7 3) A square ft 14 ft?

91 Slide 91 / 240 Try this... Find the length of the missing sides of the right triangle o y 60o x

92 Slide 92 / ft 30o Try this... Find the area of the triangle.

93 Slide 93 / Find the value of x. B 7 3 C (7 2)/2 D 14 60o 7 30o x A 7

94 Slide 94 / Find the value of x. B 7 3 C (7 2)/2 D 14 x 7 2 A 7

95 Slide 95 / Find the value of x. B 7 3 C (7 2)/2 D 14 30o x 60o A 7 7 3

96 Slide 96 / The hypotenuse of a 30o -60o -90o triangle is 13 cm. What is the length of the shorter leg?

97 Slide 97 / The length the longer leg of a 30o -60o -90o triangle is 7 cm. What is the length of the hypotenuse?

98 Slide 98 / 240 Real World Example The wheelchair ramp at your school has a height of 2.5 feet and rises at angle of 30o. What is the length of the ramp?

99 Slide 99 / 240? 30o The triangle formed by the ramp is a 30o-60o-90o right triangle. The length of the ramp is the hypotenuse. hypotenuse = 2(shorter leg) hypotenuse = 2(2.5) hypotenuse = 5 The ramp is 5 feet long. 2.5

100 Slide 100 / o? 3 feet 36 A skateboarder constructs a ramp using plywood. The length of the plywood is 3 feet long and falls at an angle of 45. What is the height of the ramp? Round to the nearest hundredth.

101 Slide 101 / o 3 feet? 37 What is the length of the base of the ramp? Round to the nearest hundredth.

102 Slide 102 / The yield sign is shaped like an equilateral triangle. Find the length of the altitude. 20 inches

103 Slide 103 / The yield sign is shaped like an equilateral triangle. Find the area of the sign. 20 inches

104 Slide 104 / 240 Trigonometric Ratios Return to the Table of Contents

105 Slide 105 / 240 Right triangle trigonometry is the study of the relationships between the sides and angles of right triangles. A c b C a B

106 Slide 106 / 240 Ever since the construction of the Bell Tower in the 1100's, it has slowly tilted south and is at risk of falling over. If the angle of slant ever fall's below 83 degrees, it is feared the tower will collapse. Leaning Tower of Pisa, Bell Tower in Pisa, Italy

107 Slide 107 / 240 Engineers can measure the angle of slant using any of the right triangles constructed below. ABC~ DBE~ FBG A D F WHY? BGEC angle of slant Engineers very carefully measure the perpendicular distance from a tower window (points A, D or F) to the ground (points G, E or C). Then they measure the distance from the tower to points C, E or G.

108 Slide 108 / 240 Triangle Height Base Ratio Height / Base ABC AC=50m BC=5m 50/5=10 DBE DE=30m BE=3m 30/3=10 FBG FG=20m BG=2m 20/2=10 Notice that all of the ratios are the same. WHY? The ratio of height/base is also called the slope ratio (rise/run) or tangent ratio. Let's calculate the ratio's of the height to the base for each right triangle.

109 Slide 109 / 240 When the triangle is dilated (pull scale), how does the angle change? What happens to the slope ratio? What happens to the ratio when the angle increases? What happens to the ratio when the angle decreases? Click for interactive website to investigate.

110 Slide 110 / 240 To learn right triangle trigonometry, first you need to be able to identify the sides of a right triangle. In a right triangle, there are 2 acute angles. In the triangle to the left, A and B are the acute angles. A Label the sides of a triangle with the lower case letter of the opposite angle. c b C a B

111 Slide 111 / 240 A Let's look at A, when A is the reference angle, the side opposite A is a. the side adjacent (or next to) A is b. b and the hypotenuse is c. adj A C c b opp C adj a hyp B c hyp a opp When B is the reference angle, the side opposite B is b. the side adjacent (or next to) B is a. and the hypotenuse is c. B

112 Slide 112 / What is the side opposite to J? B LK J L C KJ K A JL

113 Slide 113 / What is the hypotenuse of the triangle? J L B LK C KJ K A JL

114 Slide 114 / What is the side adjacent to J? A JL J L C KJ K B LK

115 Slide 115 / What is the side opposite K? J L B LK C KJ K A JL

116 Slide 116 / What is the side adjacent to K? J L B LK C KJ K A JL

117 Slide 117 / 240 Trigonometric Ratios A trigonometric ratio is the ratio of the two sides of a right triangle. There are 3 ratios for each acute angle of a right triangle. The ratios are called sine, cosine, and tangent (abbreviated sin, cos, and tan). A c b C a B

118 Slide 118 / 240 The 3 Trigonometric Ratios sinθ = opposite side hypotenuse cosθ = adjacent side hypotenuse opposite side adjacent side tanθ = A This spells... SOHCAHTOA or which is a pneumonic to help you remember the sides of a right triangle (you'll need to remember the spelling). θ c b C a B

119 Slide 119 / 240 Click for a SOHCAHTOA song on youtube.com "Gettin' Triggy Wit It".

120 Slide 120 / 240 D Example Find the sin F, cos F, and tan F E F 8 Since F is your reference angle, label the sides of the triangle opposite, adjacent and hypotenuse. Use the pneumonic to find the trig ratios. Always reduce D opp 10 6 E adj 8 opp 6 fractions to lowest terms. 3 sinf = hyp = 10 = 5 hyp F cosf = adj = 8 = 4 tan F hyp 10 = opp = 6 8 adj 5 = 3 4

121 Slide 121 / 240 D Example Find the sin D, cos D, and tan D E F 8 Since D is your reference angle, label the sides of the triangle opposite, adjacent and hypotenuse. Use the pneumonic to find the trig ratios. Always reduce D adj 10 6 E opp 8 opp 8 fractions to lowest terms. 4 sind = hyp = 10 = 5 hyp F cosd = adj = 6 = 3 hyp tand = 10 5 opp = 8 = adj

122 Slide 122 / What is the sin R? B 21/20 C 21/29 D 20/21 A 20/29

123 Slide 123 / What is the cosr? B 21/20 C 21/29 D 20/21 A 20/29

124 Slide 124 / What is the tanr? B 21/20 C 20/29 D 21/29 A 20/21

125 Slide 125 / What is the sinq? B 21/20 C 21/29 D 29/20 A 20/29

126 Slide 126 / 240 A 20/29 B 21/20 C 21/29 D 29/21 49 What is the cosq?

127 Slide 127 / What is the tanq? B 21/20 C 21/29 D 20/21 A 20/29

128 Slide 128 / 240 The angle of slant of the Tower of Pisa is 84.3 To find the trigonometric ratio of an angle, use a calculator or a trig table. Check that your calculator is set for degrees (not radians) and round your answer to the ten thousandth place (4 decimal places). Find the following: click sin 84.3 =.9951 cos 84.3 =.0993 click tan 84.3 = click A D F B angle of slant C

129 Slide 129 / 240 A 0.5 B C D Evaluate sin 60. Round to the nearest ten thousandth.

130 Slide 130 / Evaluate cos 60. Round to the nearest ten thousandth. B C D A 0.5

131 Slide 131 / Evaluate tan 60. Round to the nearest ten thousandth. B C D A 0.5

132 Slide 132 / 240 Trig tables were used by early mathematicians and astronomers to calculate distances that they were unable to measure directly. Today, calculators are usually used. x

133 Slide 133 / 240 How do you find an unknown side measure in a right triangle when you are given an acute angle and one side? You need to identify the correct trig function that will find the missing side. Use SOHCAHTOA to help. B is your angle of reference. Label the given and unknown sides of your triangle opp, adj, or hyp. Identify the trig funtion that uses B, the unknown side and the given side. Using, I am looking for o and I have a, so the ratio is o/a which is tangent. now you can solve for x, the missing side. opp adj

134 Slide 134 / 240 Example Find the trig equation that will find x. x C 30o 12 A B

135 Slide 135 / 240 Example Find the trig equation that will find x. B C 12 A x 30o

136 Slide 136 / 240 Example Find the trig equation that will find x. 30o C 12 x A B

137 Slide 137 / Using B, which is the correct trig equation needed to solve for x. A sin40 = 12/x o 0 4 B cos40 = x/12 C tan40 = 12/x x 12 B D sin40 = x/12 E D

138 Slide 138 / Using D, which is the correct trig equation needed to solve for x. B B cos50 = x/12 C tan50 = 12/x x D sin50 = x/12 E A sin50 = 12/x 12 50o D

139 Slide 139 / 240 A tan32 = x/11 11 B cos32 = x/11 C tan32 = 11/x D sin32 = 11/x K J 32o x L 56 Using J, which is the correct trig equation needed to solve for x.

140 Slide 140 / o A tan58 = x/11 B cos58 = x/11 C tan58 = 11/x D sin 58 = 11/x J x K 11 L 57 Using K, which is the correct trig equation needed to solve for x.

141 Slide 141 / 240 Finding the Missing Side of a Right Triangle Now, you can solve for x, the missing side. Round your answer to the nearest tenth. Using your calculator, find the tan 84.3 Round your answer to 4 decimal places. opp You can rewrite with a denominator of 1 and use the cross product property or multiply both sides of the equation by 5 using the multiplication property of equality (see next slide). adj

142 Slide 142 / 240 Finding the Missing Side of a Right Triangle Now, you can solve for x, the missing side. Round your answer to the nearest tenth. opp Multiply both sides of the equation by 5 using the multiplication property of equality. adj

143 Slide 143 / 240 Example Find x. Round your answer to the nearest hundredth. 25 o E x 12 M G

144 Slide 144 / 240 Example Find x. Round your answer to the nearest hundredth. x 12 o 65 M E G

145 Slide 145 / 240 Example Find y. Round your answer to the nearest hundredth. C 10 20o E y A

146 Slide 146 / 240 P 12 L 68o M 58 Find the length of LM. Round your answer to the nearest tenth.

147 Slide 147 / 240 P 12 L 68o M 59 Find the length of LP. Round your answer to the nearest tenth.

148 Slide 148 / 240 Explain and use the relationship between the sine and cosine of complementary angles.

149 Slide 149 / 240 Find the measure of A?

150 Slide 150 / 240 To find the measure of A... The sum of the interior angles of any triangle is equal to 180 degrees. A and B are complementary angles. Complementary angles are two angles whose sum of their measures is 90 degrees. The acute angles of a right triangle are always complementary.

151 Slide 151 / For right triangle ABC, what is the measure of B? B A 30 degrees B 50 degrees C 60 degrees D cannot be determined A 30o C

152 Slide 152 / If the, find the complementary angle? B 70 degrees C 160 degrees D none of the above A 20 degrees

153 Slide 153 / 240 B Let's compare the sine and cosine of the acute angles of a right triangle. In a right triangle, the acute angles are complementary. m A + m B = = sin A = 4/5 sin 53.1 =.7997 cos B = 4/5 cos 36.9 =.7997 sin A = cos B 53.1 sin 53.1 = cos 36.9 C A 3 The sine of an angle is equal to the cosine of its complement. cos A = 3/5 cos 53.1 =.6004 sin B = 3/5 sin 36.9 =.6004 cos A = sin B cos 53.1 = sin 36.9 The cosine of an angle is equal to the sine of its complement.

154 Slide 154 / 240 First, find the measure of LP using the sine function. Then, find the measure of LP using the cosine function. sine function cosine function x L 68o Sine and Cosine are called co-functions of each other. Co-functions of complementary angles are equal. P 22o 12 M

155 Slide 155 / Given that sin 10 =.1736, write the cosine of A sin 10 =.1736 B sin 80 =.9848 C cos 10 =.9848 D cos 80 =.1736 a complementary angle.

156 Slide 156 / Given that cos 50 =.6428, write the sine of A sin 50 =.7660 B sin 40 =.6428 C cos 50 =.6428 D cos 40 =.7660 a complementary angle.

157 Slide 157 / Given that cos 65 =.4226, write the sine of a complementary angle. B cos 25 =.9063 C sin 65 =.9063 D cos 65 =.4226 A sin 25 =.4226

158 Slide 158 / What can you conclude about the sine and cosine of 45 degrees? Students type their answers here

159 Slide 159 / 240 Solving Right Triangles Return to the Table of Contents

160 Slide 160 / 240 To solve a right triangle means to find all 6 values in a triangle. The lengths of all 3 sides and the measures of all 3 angles. A c b C a B

161 Slide 161 / 240 Let's solve a right triangle given the length of one side and the measure of one acute an gle (AAS). You need to find the 3 missing parts. A x 15 C 64o y z B

162 Slide 162 / 240 First, let's find the measure of A. A x 15 z B C 64o y

163 Slide 163 / 240 Then, let's find the measure of AB. A z B C 64o y

164 Slide 164 / 240 Then, let's find the measure of BC. A z B C 64o 13.48

165 Slide 165 / 240 Try this... Find the missing parts of the triangle. 11 E 37o D R

166 Slide 166 / 240 Let's solve a right triangle given the length of two sides (SSA). A 9 B x z 15 y C

167 Slide 167 / 240 First, find the length of BC since we know how to do that. But, how do you find the measure of A and C? 9 B x z 15 y C A

168 Slide 168 / 240 You will need to use the inverse trig functions. If sinθ =, θ = sin If cosθ =, θ = cos-1 If tanθ =, θ = tan-1 A -1 c b C Pronounced inverse sine, inverse cosine, and inverse tangent. a θ B With the sine, cosine and tangent trig functions, if you know the angle θ and the measure of one leg, then you can find the measure of a leg of a triangle. With the inverse sine, inverse cosine and inverse tangent trig functions, if you know the measures of 2 legs of a triangle, you can find the measure of the angle.

169 Slide 169 / 240 The 3 Inverse Trigonometric Ratios θ = sin-1( opposite side ) hypotenuse θ = tan-1( opposite side ) θ = cos-1( adjacent side ) adjacent side hypotenuse Use the inverse trig function to find the unknown angle measure when you know the length of 2 sides. A Remember: c b C a θ B

170 Slide 170 / Find sin Round the angle measure to the nearest hundredth.

171 Slide 171 / Find tan Round the angle measure to the nearest hundredth.

172 Slide 172 / Find cos Round the angle measure to the nearest hundredth.

173 Slide 173 / 240 To find an unknown angle measure in a right triangle, You need to identify the correct trig functionthat will find the missing value. Use SOHCAHTOA to help. Using cosine. 9 adj θ 15 hyp, I have a and h, so the ratio is a/h which is now you can solve for A, the missing angle using the inverse trig function. How are you going to calculate the measure of C? B A is your angle of reference. Label the two given sides of your triangle opp, adj, or hyp. Identify the trig funtion that uses A, and the two sides. A C

174 Slide 174 / 240 Example Find the trig equation that will find θ. B 7 C 12 A θ

175 Slide 175 / 240 Example Find the trig equation that will find θ. 10 C θ 12 A B

176 Slide 176 / 240 Example Find the trig equation that will find θ. θ C 12 9 A B

177 Slide 177 / Which is the correct trig equation to solve for B A 7 12 B C D E D

178 Slide 178 / Which is the correct trig equation to solve for B B 5 C D E 12 A D

179 Slide 179 / 240 A K B 11 C D J 9 L 71 Which is the correct trig equation to solve for

180 Slide 180 / 240 Try this... Solve the right triangle. Round your answers to the nearest hundredth. R Q 7 S 24

181 Slide 181 / Find CE. E D 5 8 C

182 Slide 182 / Find m C. E D 5 8 C

183 Slide 183 / Find the m E. E D 5 8 C

184 Slide 184 / Find the m G. o 20 L A 18 G

185 Slide 185 / 240 o Find AL. L A 18 G

186 Slide 186 / Find the m P. P A 49.19o 12 C 41.81o D 56.31o E 18 N B 33.69o

187 Slide 187 / Find RT. S T B C D o R A 10.44

188 Slide 188 / 240 Angle of Elevation and Depression Return to the Table of Contents

189 Slide 189 / 240 How can you use trigonometric ratios to solve word problems involving angles of elevation and depression?

190 Slide 190 / 240 When you look up at an object, the angle your line of sight makes with a line drawn horizontally is the angle of elevation.

191 Slide 191 / 240 When you look down at an object, the angle your line of sight makes with a line drawn horizontally is the angle of depression.

192 Slide 192 / 240 The angle of elevation and the angle of depression are both measured relative to parallel horizontal lines, they are equal in meaure.

193 Slide 193 / How can you describe the angle relationship between the angle of elevation and the angle of depression? C alternate exterior angles D none of the above A corresponding angles B alternate interior angles

194 Slide 194 / 240 Example Amy is flying a kite at an angle of 58o. The kite's string is 158 feet long and Amy's arm is 3 feet off the ground. How high is the kite off the ground? 15 x 8f ee t 58o 3 feet

195 Slide 195 / 240 x 15 8f t 58o sinθ = x 158 sin58 = x = x 158 x = 134 Now, we must add in Amy's arm height = 137 The kite is about 137 feet off the ground.

196 Slide 196 / 240 Example You are standing on a mountain that is 5306 feet high. You look down at your campsite at angle of 30o. If you are 6 feet tall, how far is the base of the mountain from the campsite? 30o 6 ft 5306 ft x

197 Slide 197 / 240 tan30 = 5312 ft 30o x.5774 = 5312 x 5312 x.5774x = 5312 x 9,200 ft The campsite is about 9,200 ft from the base of the mountain.

198 Slide 198 / 240 Try this... You are looking at the top of a tree. The angle ofelevation is 55o. The distance from the top of the tree to your position (line of sight) is 84 feet. If you are 5.5 feet tall, how far are you from the base of the tree?

199 Slide 199 / 240 A elevation B depression 80 When you look down at an object, the angle your line of sight makes with a line drawn horizontally is the angle of.

200 Slide 200 / Katherine looks down out of the crown of the statue of liberty to an incoming ferry about 345 feet. The distance from crown to the ground is about 250 feet. What is the angle of depression?

201 Slide 201 / What is the distance from the ferry to the base of the statue?

202 Slide 202 / 240 Law of Sines and Law of Cosines Return to the Table of Contents

203 Slide 203 / 240 How can you solve a non-right triangle? How can you find the side lengths and angle measures of non-right triangles? The Law of Sines and Law of Cosines can be used to solve any triangle. You can use the Law of Sines when you are given 1. Two angle measures and any side length (AAS or ASA) 2. Two side lengths and the measure of a non-included angle (SSA) when the angle is a right angle or an obtuse angle. The Law of Sines has a problem dealing with SSA when the angle is acute. There can be zero, one or two solutions. Click on: Khan Academy Video "More On Why SSA Is Not A Postulate" for more info. You can use the Law of Cosines when you are given 3. Three side lengths (SSS) 4. Two side lengths and the measure of an included angle (SAS)

204 Slide 204 / 240 C Law of Sines b A a c If ABC has sides of length a, b, and c, then sin A = sinb = sinc a b c To use the Law of Sines, 2 angles and 1 side must be given. B

205 Slide 205 / 240 C Let's prove the Law of Sines If ABC has sides of length a, b, and c, then sin A = sinb = sinc A a b c Given: ABC has sides of length a, b, and c Prove: sin A = sinb = sinc a b c b h c a B Statements Reasons ABC with side lengths a, b, and c click Given Def of Altitude Draw an altitude from C to side AB click Let h be the length of the altitude Def of sine click Multiply click by b. Mult Prop of =. Multiply click by a. Mult Prop of =. Substitution Prop of = click Divide by ab. Division Prop of = click

206 Slide 206 / 240 C Prove the Law of Sines (continued) Given: ABC has sides of length a, b, and c Prove: sin A = sinb = sinc a b c b A h c C a b g B A Statements a c B Reasons Def of Altitude Draw an altitude from B to side AC click Let g be the length of the altitude Def clickof sine Multiply click by c. Mult Prop of =. Multiply click by a. Mult Prop of =. Substitution Prop of = click Divide by ac. Division Prop of = click Substitution Prop of = click

207 Slide 207 / 240 Use the Law of Sines to solve the triangle. sin A = sinb = sinc a b c Select the ratios based on the given information. B C 70o 65o b Given: m B, m C and BA (side c) (AAS) 10 A Which ratios must be used first? Why? a

208 Slide 208 / 240 C sinb = sinc b c sin70 = sin65 b 10 a 70o 65o b 10 A First we can find the length side b. B

209 Slide 209 / 240 C B a 70o 65o Triangle Sum Theorem m A + m B + m C = 180o b= A Before we find the length of side a, we find the m A.

210 Slide 210 / 240 B a C sina = sinc a c 65o 70o 10 =45o b=10.37 A Now we find the length side a.

211 Slide 211 / 240 Try this... Use the Law of Sines to find the length of side b (ASA). C 85o 9 29o A b a B Since the length of the side opposite <C is given, find the m<c first. hint

212 Slide 212 / 240 Example... Find the length of side b (SSA with an obtuse angle). 8 A 101o b 2.8 C B

213 Slide 213 / Find the m A. B 31o C 29o D 28o 70o c A 10 81o C b A 19o B

214 Slide 214 / Which ratio must be used to find the length of B b or c? 70o c 81o A sina a B sina b C b C sinb b D sinc c A 10

215 Slide 215 / What is the length of b? B c A 10 81o C b 70o

216 Slide 216 / What is the length of c? B c A 10 81o C b 70o

217 Slide 217 / 240 B Law of Cosines c A a b C If ABC has sides of length a, b, and c, then: To use the Law Cosines, you must be given the length of 3 sides (SSS) or the length of 2 sides and the measure of the included angle (SAS).

218 Slide 218 / 240 Let's prove the Law of Cosines A Given: ABC has sides of length a, b, and c Prove: (similar reasoning shows that ) b a b If ABC has sides of length a, b, and c, then C C c B A Statements h x a D c-x c B Reasons ABC with side lengths a, b, and c Given click Def of Altitude Draw an altitude CD from C to side click AB. Let h be the length of the alt. Let x be the length of AD. Then (c-x) is the length of DB. Segment Addition Postulate click In Definition of cosine click ADC, cosa = x/b (1) x=b(cosa) Multiply click by b. Mult Prop of =. (2) In Pythagorean Theorem click In ADC, CDB, Pythagorean Theorem click Simplify click Substitution, equation (2) click Associative Prop of Addition click Substitution, equation (1) click

219 Slide 219 / 240 Example Use the Law of Cosines to solve the right triangle. B a=16 C a is opposite <A b is opposite <B c is opposite <C c=27 b=23 A The formula you choose depends on which angle you are solving for first.

220 Slide 220 / 240 B c=27 a=16 To find the m A, C b=23 a2 = b2 + c2-2bc(cosa) 162 = (23)(27)(cosA) 256 = (cosA) 256 = (cosA) = -1242(cosA).8068 = cosa A = cos-1(.8068) m A 36.22o A

221 Slide 221 / 240 B a=16 c= To find the m B, C b2 = a2 + c2-2ac(cosb) 232 = (16)(27)(cosB) 529 = (cosB) 579 = (cosB) -406 = -864(cosB).4699 = cosb B=cos-1(.4699) m B 61.97o b=23 A or Using 2 different methods, the answers are slightly different because of rounding.

222 Slide 222 / 240 B C c= b=23 To find the m C, Use the Triangle Sum Theorem. A a=16

223 Slide 223 / 240 Try this... Use the Law of Cosines to find the m<b (SSS). 7 A 6 5 B C

224 Slide 224 / In the triangle the length of c is... B 9 C 15 A 15 9 C 8 B A 8

225 Slide 225 / In the triangle the length of a is... B 9 C 15 A 15 9 C 8 B A 8

226 Slide 226 / Which formula would you use to find the m<a? B a2 = b2 + c2-2bc(cosa) C b2 = a2 + c2-2ac(cosb) D c2 = a2 + b2-2ab(cosc) A a2 = b2 + c2-2ac(cosa)

227 Slide 227 / What is the m A? A 15 C 8 B 9

228 Slide 228 / What is the m C? A 9 C 8 B 15

229 Slide 229 / What is the measure of B (ASA)? Students type their answers here B C A

230 Slide 230 / The Law of Sines and Cosines is used to solve... B acute triangles C obtuse triangles D all triangles A right triangles

231 Slide 231 / 240 Area of an Oblique Triangle Return to the Table of Contents

232 Slide 232 / 240 Do you remember this? Previously, we found the area of a triangle when we were given 3 sides. Find the area of the triangle. 13 feet 13 feet 10 feet

233 Slide 233 / 240 A = 1 2 bh b is the base of the triangle b = 10. h is the altitude (or height). It is the perpendicular bisector of the base in an isosceles triangle. Find h, using the pythagorean theorem - 13 feet h 5 feet 13 feet 5 feet

234 Slide 234 / 240 What formula can you use to find the area of a triangle if you know the length of two sides and the measure of an included angle (SAS)? Find the area of the triangle. 13 feet feet

235 Slide 235 / 240 Since A = 1 bh 2 and b = 10, we need to find h. 13 feet h feet

236 Slide 236 / 240 Let's derive the formula for an oblique triangle. Given: ABC has sides of lengtha, b, and c. Altitude h. Prove: B a C h Statements Reasons click ABC with side lengths a, b, and c Given Draw an altitude from B to side AC Def clickof Altitude c Let h be the length of the altitude b A Def clickof sine Multiply click by a. Mult Prop of =. Definition. Formula for the area of a triangle. click Substitution Prop of = click Commutative Prop of Multiplication click

237 Slide 237 / 240 A D B E C F 94 Which of the following expressions can be used to find the area of the triangle below? Select all that apply.

238 Slide 238 / Find the area of the triangle to the nearest tenth. Students type their answers here

239 Slide 239 / Find the area of the triangle to the nearest tenth. Students type their answers here

240 Slide 240 / Find the area of the triangle to the nearest tenth. Students type their answers here

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