( y) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios, Mixed Exercise 9. 2 b. Using the sine rule. a Using area of ABC = sin x sin80. So 10 = 24sinθ.

Size: px

Start display at page:

Download "( y) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios, Mixed Exercise 9. 2 b. Using the sine rule. a Using area of ABC = sin x sin80. So 10 = 24sinθ."

Bruno Stevens
6 years ago
Views:

1 Trigonometric ratios, Mixed Exercise 9 b a Using area of ABC acsin B 0cm 6 8 sinθ cm So 0 4sinθ So sinθ 0 4 θ 4.6 or 3 s.f. (.) As θ is obtuse, ABC 3 s.f b Using the cosine rule b a + c ac cos B AC cos B 87.6 AC 3.68 The third side has length 3.7 m (3 s.f.). a c Using the sine rule sin x sin80 6 sin 80 sin x x. 3 s.f. The angle between the cm and 6 cm sides 80 + x 00 x. is 80 Using the area of a triangle formula: area 6 sin 00 x cm 0.6 cm 3 ( s. f. ) Using the cosine rule cos x x cos (.) x s.f Using the area of a triangle formula area. 3 sin x cm (.).37 cm 3 s.f Use the sine rule to find the angle opposite the 3 cm side. Call this y. sin y sin sin 40 sin y y.68 ( y) So x s.f. Area of triangle 3 sin x 66.6 cm 3 ( s. f. ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

2 3 4 b 4 a Use the cosine rule to find angle A cos A 3 0. cos 0. A 0 Area of triangle 3 sin Acm 6.49 cm (.f.) 6.0 cm 3 s sin ADB sin sin 7 sin ADB sin 7 ADB sin ( ADB) So ABD Area of ABD sin ABD 7.04 cm In BDC, BDC 80 ADB 8.9 Area of BDC sin BDC 4.0 cm Total area are a ABD + area BDC ( f. ).0 cm 3 s. BD In BDA, sin So BD 8.sin AD cos AD 8. cos ABD We can use AD and BD to calculate the area of ABD or use: Area of ABD 8. BD sin cm Area of BDC 0.4 BD sin00.37 cm Total area are a ABD + area BDC ( f. ) 36.cm 3 s. a Using the cosine rule: b a + c ac cos B a ( ) ( a ) + a a ( a )( a ) a a 3 0 cos0 a + + a as a> 0 a Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

3 b Area ABC 0 3 sin cm Using the area formula: sinθ sinθ θ 4 or 3 But as θ is not the largest angle, θ must be 4. Use the cosine rule to find x. x x + cos So x So the triangle is isosceles with two angles of 4. It is a right-angled isosceles triangle. 7 a AB ( 3 0) + ( 4 ) 8 8 c Using the cosine rule a + b c cosc ab + 8 cosc Find sin C by using the identity, cos x+ sin x or by drawing a 3,4, triangle and looking at the ratio of the sides. b Using the area formula: area of ABC absin C sin C. cm 7 a Use Pythagoras theorem. a ( 0) ( 3 ) AC + b ( 3 ) ( 4 3) BC + a Using the cosine rule ( x ) ( x+ ) + ( x ) ( x )( x ) 4 4 x x+ x + x x x+ x + x 4x 0 x x cos0 + ( x x+ ) + ( x ) x 4 x> b Area of ( x ) ( x ) + sin0 3 sin0 3 3 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 3 8 b Area of cm (3 s.f.) 9 a 0 b a + c ac cos B cos So b So point C is.0 km from the park keeper s hut. b sin A sin B a b sin A sin sin 70 sin A. So A Bearing 360 ( ) 4.9 The bearing of the hut from point C is 4. c Area of acsin B.4. sin km (3 s.f.) Using triangle ABD, the angles are, 48 and 7. b d sin B sin D b 7 sin48 sin7 7sin48 b sin7 b Using the larger right-angled triangle: height sin height sin The height of the church tower is 3. m (3 s.f.). a A stretch of scale factor in the x direction. b A translation of +3 in the y direction. 3 a c A reflection in the x-axis. d A translation of 0 in the negative x direction (i.e. 0 to the left). b + c a cos A bc 0 + cos A (0)() cos A 600 So A 36.7 Area of one sail bcsina 0 sin Area of all four sails 39 m (3 s.f.) b tan (x 4 ) + cos x 0 tan (x 4 ) cos x The graphs do not intersect so there are no solutions. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 4 a As it is the graph of y sin x translated, the gap between A and B is 80, so p 300. b The difference in the x-coordinates of D and A is 90, so the x-coordinate of D is 30. The maximum value of y is, so D is the point (30, ). 6 b So sin α sin (80 α), sin(80 \+ α) and sin(360 α) have the same y value, which will be k. So sinα sin(80 α) sin(80 + α) sin(360 α) 7 a c For the graph of y sin x, the first positive intersection with the x-axis would occur at 80. The point A is at 0 and so the curve has been translated by 60 to the left. k 60 d The equation of the curve is y sin (x + 60). 3 3 When x 0, y sin 60, so q. a The graph of y sin x crosses the x-axis at (80, 0). f(x) sin px is a stretch horizontally with scale factor f(x) sin x p b i From the graph of y cos θ, which shows four congruent shaded regions, if the y value at α is k, then y at (80 α) is k, y at (80 + α) is k and y at (360 α) it is +k. So cosα cos (80 α) cos (80 + α) cos (360 α) 6 a b The period of f(x) is ii From the graph of y tan θ, if the y value at α is k, then at (80 α) it is is k, at (80 + α) it is +k and at (360 α) it is k. So tan α tan (80 α) +tan (80 + α) tan (360 α) 8 a b The four shaded regions are congruent therefore the magnitude of the y value is the same for sin α. Sin α and sin (08 α) have the same y value (call it k). b There are 4 complete waves in the interval 0 x 4 so there are 4 sand dunes in this model. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

6 8 c The sand dunes may not all be the same height. Challenge ACB tan 4 Show that θ + ϕ 4 sin θ Using the sine rule: sin (80 θ φ) sin θ 0 sin (80 θ φ) sin 0θ Substituting sin θ sin (80 θ φ) : 0 0 sin 4, but angle 80 θ ϕ is obtuse. So, 80 θ ϕ Therefore, θ + ϕ 4 So, AEB + ADB ACB Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios 9E. b Using the line of symmetry through A. 1 a. cos 48 = 14.6 So y = 29.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios 9E. b Using the line of symmetry through A. 1 a. cos 48 = 14.6 So y = 29. Trigonometric ratios 9E a b Using the line of symmetry through A y cos.6 So y 9. cos 9. s.f. Using sin x sin 6..7.sin 6 sin x.7.sin 6 x sin.7 7.6 x 7.7 s.f. So y 0 6+ 7.7 6. y 6. s.f. b a Using sin sin