( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios 9E. b Using the line of symmetry through A. 1 a. cos 48 = 14.6 So y = 29.
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1 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 x 7.7 s.f. So y y 6. s.f. b a Using sin sin A z.7 sin y sin 6.7 sin y So z sin s.f. b x 0 ( + ) x c d x Using the line of symmetry through C 0 cos0 y 0 So y cos0. s.f. Since AC is isosceles with AC C z. s.f.. As angle angle C, z.6 cm. sin A sin Using a b sin0 sin x. 6 6sin0 So sin x x.0( s.f. ) ( x ) So y Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
2 d y 9.0 ( s.f. ) Using z. sin y sin0.sin y So z sin0.09 s.f. f e Using the line of symmetry through 6 cos y y cos.0 y. ( s.f. ) As the triangle is isosceles a + b c Using ab + 6 cos x 0.06 ( f. ) x 9. s. sin sin C Using b c sin y sin x 6 sin x sin y 6 sin x y sin y 6. s.f. Using the angle sum for a triangle z 0 x+ y 9.96 z 9.9 s.f. z y. s.f. So x 0 y+ z 97. x 97. s.f. g sin A sin C Using a c sin x sin sin 0 sin x 9. 0.sin 0 x sin or 9. 0.sin 0 x 0 sin 9. x.7 or x.7 x x. s.f. or s.f. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
3 b c g Using sin sin C z 9. sin y sin 0 9.sin y z sin 0 When x. y So y 9.7 s.f. 9.sin y z sin 0.7 s.f. When x.7 y So y.7 s.f. 9.sin y z sin 0.6 s.f. So x., y 9.7, z.7 or x, y.7, z.6 h a b + c bc A Using cos x x 7.07 s.f. Using sin y sin 9.6 x 9.6sin sin y x 9.6sin y sin x. 9.6 cos h y 7.6 i y 7.7 ( s.f. ) Then z 0 ( ) 6. z 6. ( s.f. ) So x 7.07, y 7.7, z 6. a + c b Using cos ac cos x x 9.77 x 9. s. f. In right-angled triangle AD y sin x. So y.sin x 9.9 s.f. In right-angled triangle ACD y sin z z 7.0 z 7.0 ( s.f. ) So x 9., y 9.9, z 7.0 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
4 a b Using sin C sin sin 0 sin C 0..sin 0 C sin 0... ( s.f. ) 0 ( 0 + C) s.f. b a Using sin sin A 0.sin b sin 0.cm ( s.f. ) Area acsin 0.. sin0.. cm ( s.f.) b 6.sin0 sin A A 0.66 A.9. ( s.f. ) So 0 ( 0 + A). ( s.f. ) Area acsin 6..9 sin0.6.7 cm ( s.f.) a Angle AC As A C, all angles are 60. It is an equilateral triangle. So AC km. b As AC 60, the bearing of C from A is 060. Using + cos b ab C A A 9.6 (.. f ) 9.6 cm s sin A sin C Using a c 6..9 cos0 From the diagram AC Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
5 b c Using sin sin C AC 70 sin0 sin 0 70sin0 AC sin AC 07 km s.f. From the diagram, the bearing of C from A is 0. Using the sine rule sin C sin sin 0 sin C 0 0. C C sin 0. or 0 sin 0. C. or C 6.9 AC., AC 7 s.f. (Store the correct values; these are not required answers.) Triangle CC is isosceles, so CC can be found using this triangle, without finding AC and AC. Use the line of symmetry through CC cos CC 0 CC 0cos CC 0cos AC 0cos. CC km 6 a b In the isosceles DC DC So DA Using the sine rule in AD sin A sin D a d sin A sin00 sin00 sin A sin00 So A sin 7.96 AD A.0 b d Using sin sin D x sin sin00 sin x sin00.6 x. s.f. c + d a In ADC, using cos A cd Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
6 6 b c cos A So A. Using the sine rule in AC a b sin A sin x 0 So sin A sin 60 0sin A x sin 60 So x 7.9 s.f. 7 a As A + C > AC + x + > 7 x + > x > As A + AC > C + 7> x + 9 > x So < x < 9 In AC, c cm, b. cm, AC 0, (0 0) sin sin C Using b c.sin0 sin So A 0 ( 0 + ).6 In ADC, c. cm, d. cm, A.6 Using the cosine rule: a c + d cd cos A So DC cos A.99 DC.79 cm Using the sine rule:.sin A sin x DC x 6. ( s.f.) b Using b a + c ac cos 7 x + + x + cos60 i 9 x + x+ + 6 ( x+ ) x + x+ + x So x 7 x 6.0 ( s.f. ) Area acsin.0 sin cm ( s.f.) 9 6 ii 7 (x + ) + (x + ) cos 9 x + x+ + 6 ( cos ) x 6cos So: x + cos x 9 + 6cos 0 + ( ) or x x Use the quadratic formula b ± b ac x with a a b cos.66 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
7 7 b ii c ( 9 + 6cos ) a 9 ( 9 ) + 0. x 7. s.f. (The other value of x is less than.) Area acsin 9. sin.0.cm ( s.f.) Using b a + c ac cos where cos ( x + ) x + x+ 6 + x + x 0 ( x+ 6)( x ) 0 So x ( x> ) b Use identity, cos x+ sin x. cos 9 so sin AC Area acsin cm ( s.f.) 9 Using sin 60 sin C sin 60 C sin b a Using sin sin A AC sin 7 sin 60 sin 7 So AC sin 60.9 cm s.f. 0 a Using the cosine rule: b a + c ac cos ( ) ( ) ( x )( x) AC x + + x ( ) ( ) ( x )( x) AC x x + cos0 AC x + x + + x + x + + AC x + x + + x + x x + x x+ + 7 b Completing the square: ( x ) x x x 6 + This is a minimum when x 0 x. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
8 a b a Using sin sin A AC sin 0 sinθ sin 0 AC ( ) sin 0 AC ( )( sin 0 ) 0 cm Using the cosine rule a b + c bc cos A a x, b x, c 7and A 60 with x ( x) + 9 ( x) 7 x 6 6x+ x + 9 7( x) x 6 6x+ x x 9x 7 x So C 6 cm and AC 6 cm Area cm bcsin A 7 sin cm ( s.f.) cos 60 a a + b c ab + ()() C C 6. ( s.f.) b Use the formula. Area absin C sin cm ( s.f.) b + c a cos A bc cos A (.)(.) cos A 7.6 A 6... Angle DA 6. ( d.p.) a + b c ab (.)(7.) C Angle CD 0. b Area AD bcsin A.. sin Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
9 b Area CD absin C. 7. sin Total area The area of the flower bed is. m. c First find angle AD: a + b d cos D ab cos D (.9)(.) cos D.7 So D Now find angle DC: b + c d cos D bc cos D (.)(.9) cos D. So D Angle ADC Now find the length AC: d a + c ac cos D d cos.9 d So d. So the length of AC is. m. Use the sine rule to work out angle CED. sin E sin D e d sin E sin 0 0 0sin 0 sin E E or 06.7 The angle is obtuse so Angle CED 06.7 Angle ECD Use trigonometry to work out the height of triangle CDE. sin. height Height.7 cm The height of triangle AE cm Area of triangle Area of the shaded triangle is. cm. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9
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