Solving Right Triangles Using Trigonometry Examples

Transcription

1 Solving Right Triangles Using Trigonometry Eamples 1. To solve a triangle means to find all the missing measures of the triangle. The trigonometri ratios an be used to solve a triangle. The ratio used depends upon what measures are given and what measures are missing. Sometimes, more than one ratio an be used. 2. Review the following trigonometri ratios with students. sine of = a sin of = osine of = b os = tangent of = a tan = b measure _ of _ side _ opposite _ measure _ of _ hypotenuse measure _ of _ side _ adjaent _ to _ measure _ of _ hypotenuse measure _ of _ side _ opposite _ measure _ of _ side _ adjaent _ to _ b Make sure students are familiar with looking up trigonometri values in a table. a 3. Thought Provoker Put the following questions on the board and have the students onsider them during the lesson. fter the eamples are ompleted, have the students make omments on the questions. a. Question 1 an a right triangle be solved if the lengths of two sides are given? (YES) b. Question 2 an a right triangle be solved if the measures of both aute angles are given? (NO). Question 3 an a right triangle be solved if the length of one side and the measure of one aute angle are given? (YES) 4. Eample Find the measures of and in. eause the lengths of all sides are given, the sine, osine, or tangent ratio an be used. Suppose you use the sine ratio. sin = m 4 m Using the table or alulator, we see that the measure of is approimately 53 o. Thus, the measure of is about (90 o 53 o ), or 37 o. 3 m

2 5. Emphasize to students that they an hek their answers on problems similar to eample 4. The sum of the three angles in a triangle equal 180 degrees. If not, the student should go bak and hek work. 6. sk the students the following question. In eample 4, how would you use the osine ratio to find the measure of? os = 5 3 os = Then 53 o. 7. sk the students the following question. In eample 4, how would you use the tangent ratio to find the measure of? tan = 3 4 tan = Then 53 o Point out that when the measures of all three sides of the triangle are known, any of the three trigonometri ratios an be used to find the measures of the aute angles. 8. Eample Find the measures of and in. nswers may vary but one type of solution is: 12 sin = 13 sin = Then 67 o sin = 13 5 sin = Then 23 o 5 m 12 m 13 m F 9. Eample Find the length of DE in DEF. With respet to E, only the lengths of the adjaent side and the hypotenuse are given. In this ase, use the osine ratio. os 40 o = (0.7660)(7.6) Thus, DE is approimately 5.8 m long. D 7.6 m 40 o E m

3 10. Have students use sin 50 o = to solve for the measure of DE. (See eample above.) Do you get the same result? YES 11. EXMPLE Find the length of JK in JKL. L In this ase, use the tangent ratio. tan 48 o 5.8 = o J mm 5.8 mm K 5.2 Thus, JK is about 5.2 mm long. 12. sk the students to eplain why the tangent ratio is used in eample 11. The tangent ratio is used beause the problem involves the measures of the two legs of the triangle. 13. Eample Find the missing measures in the triangle at the right. The measure of R is (90 o 36 o ), or 54 o. sin 36 o = o y in. P 18 in Thus, QR is about 10.6 inhes long. os 36 o y = 18 y y Thus, PQ is about inhes long. Q in. R Point out that one the measures of all sides are known, the Pythagorean Theorem an be used to hek them.

4 14. Eample Find DE in DEF. F os 36 o = = D m 5.2 m E 36 o Thus, DE is about 4.2 m L 15. Eample Find JK in JKL. tan 33 o 13.1 = = J 33 o mm 13.1 mm K 20.2 Thus, JK is about 20.2 mm. 16. Eample Find the missing measures of. sin 50 o y = 8.3 y (0.7660) y os 50 o = (0.6428) = (90 50) o = 40 o y m m 8.3 m 50 o 6.4 m y 5.3 Therefore, 5.3, 6.4, = 40 o

5 Name: Date: lass: Solving Right Triangles Using Trigonometry tivity Sheet State whih trigonometri ratios you would use to find the missing measures in eah triangle o Find a 2. and b Find and y a ft 15 ft 39 o y 3. b ft 13 ft 5 ft Find and m 12 in. 15 ft Find and 9 in. Solve Eah Triangle. (When using trigonometri funtions, approimate values to 4 deimal plaes and approimate degree measure to nearest whole degree) O in. 8 ft 6 ft o 9 m m 60 o

6 ft 22 o 23 m 52 o o 16m 40 o 7 km Draw and label, and then solve eah right triangle ( is a right angle). 13. ngle = 31 o, a = 6m 14. a = 6 in., = 10 in. 15. ngle = 42 o, = 10 in. 16. b = 5 ft, a = 4 ft 17. = 14 ft, b = 11 ft 18. = 11 m, b = 6 m 19. ngle = 40 o, b = 6 m 20. ngle = 28 o, a = 16 m 21. ngle = 45 o, = 2 ft 22. ngle = 75 o, b = 3 km

7 Solving Right Triangles Using Trigonometry tivity Sheet Key State whih trigonometri ratios you would use to find the missing measures in eah triangle o 2. Sin 46 o b = 15 a ft 15 ft os 46 o a = o os 39 o 12 = y Tan 39 o = 12 b ft X 12 m Y ft 5 ft Sin = 13 5 os = in. 15 ft 12 Tan = 9 9 Tan = 12 9 in. Solve Eah Triangle O in. 8 ft = (90 21) o = 69 o os 21 o = in. tan 21 o a a = a 5 in ft 8 Tan = 53 o 6 (90 53) o 37 o sin 53 o 8 8 = ft

8 7. 70 o 9 m =( 90 70) o 20 o Sin 70 o 9 9 = m Tan 70 o 9 9 = b 3.3 m b b m 60 o = (90 60) o 30 o Sin 60 o b b = b 13.9 m os 60 o a a = m ft 22 o 23 m = (90 22) o 68 o Tan 22 o = 12 a a a 4.8 ft os 22 o = ft 52 o = (90 52) o 38 o Sin 52 o b b = b 18.1 m os 52 o a a = a 14.2 m 23 23

9 o 16m 40 o = (90 40) o 50 o Sin 40 o a a = a 10.3 m os 40 o b b = b 12.3 m = (90 35) o 55 o 7 km Sin 35 o = km Tan 35 o = b b 7 10 km Draw and label, and then solve eah right triangle ( is a right angle). (Students drawings will vary.) 13. ngle = 31 o, a = 6m = (90 31) o 59 o Tan 31 o = b b 6 b 10 m 31 o Sin 31 o = m 6 m 14. a = 6 in., = 10 in. Sin = 10 6 Sin = o (90 37) o 53 o Tan 37 o b b 6 b 8 in. 10 in. 6 in.

10 15. ngle = 42 o, = 10 in. = (90 42) o 48 o Sin 42 o b b = b 6.7 in o os 42 o = 10 a a a 7.4 in. 10 in. 16. b = 5 ft, a = 4 ft Tan = 5 4 Tan o (90 39) o 51 0 Sin 39 o ft 4 ft 5 ft 17. = 14 ft, b = 11 ft 11 os = os o 14 (90 38) o 52 0 Sin 38 o 14 a a a 8.6 ft 14 ft 11 ft 18. = 11 m, b = 6 m os = 11 6 os o (90 57) o 33 o Sin 57 o 11 a a a 9.2 m 11 m 6 m 19. ngle = 40 o, b = 6 m = (90 40) o = 50 o Sin 40 o = m 40 o Tan 40 o = a a 6 a 7.2 m 6 m

11 20. ngle = 28 o, a = 16 m = (90 28) o = 62 o Tan 28 o = 16 b b b 8.5 m 28 o os 28 o = m 16 m 21. ngle = 45 o, = 2 ft = (90 45) o = 45 o Sin 45 o a a = a 1 ft 2 2 os 45 o b b = b 1 ft ft 45 o 22. ngle = 75 o, b = 3 km = (90 75) o = 15 o Tan 75 o = 3 a a a 11.2 km os 75 o = km 75 o 3 km