F.4 Maths -- trigonometry(II)

2009-09-01 6:45 am

求mathematics in action 4b 答案 p.226 Q31(a)&(b) , Q33(b)
31.In the figure, ABCD is a cyclic quadrilateral, where AB=2cm, BC=6cm, CD=4cm and angle ABC=θ. If area triangle ABC= area of triangle ADC, find
(a) AD
(b) cosθ
33. Find the angle betweenCP and plane BCEF.

更新1:

Thank you for your answering of first question Buy you have missed the second queston Q33@@Please help me!!

回答 (1)

用戶10012

2009-09-01 6:58 am

✔ 最佳答案

Q31.
(a)Area of triangle ABC = (1/2)(AB)(BC)sin x ( x = theta)
= (1/2)(2)(6) sin x = 6 sin x.
Area of triangle ADC = (1/2)(AD)(DC) sin ( 180 - x) because ABCD is a cyclic quad.
= (1/2)(AD)(4) sin x = 2(AD) sin x
so 6 sin x = 2(AD) sin x
AD = 3.
(b) By cosine rule
AC^2 = AD^2 + DC^2 - 2(AD)(DC) cos ( 180 - x)
= 9 + 16 + 24 cos x = 25 + 24 cos x.
AC^2 = AB^2 + BC^2 - 2(AB)(BC) cos x
= 4 + 36 - 24 cos x = 40 - 24cos x
so 25 + 24 cos x = 40 - 24cos x
48 cos x = 40 - 25 = 15
so cos x = 15/48 = 5/16.

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