Maths about trigonometry

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Q1. Draw point K such that DK is perpendicular to AM.
Angle BAM + angle DAB + angle DAK = 180
so angle BAM + angle DAK = 180 - angle DAB = 180 - 90 = 90.....(1)
Angle KDA + angle DAK = 90 ....(2)
(2) = (1), so angle KDA = angle BAM = x ( instead of using theta).
KA = DA sin x = a sin x
AM = AB cos x = b cos x
so KA + AM = DN = a sin x + b cos x.



2009-08-13 07:44:48 補充：
Q7. Angle PRQ = 42 - 20 = 22 degree. Angle RPQ = 90 - 42 = 48 degree. By sine rule, RQ/sin(angle RPQ) = PQ/sin(angle PRQ). RQ/sin48 = 3/sin 22, so RQ =
3 sin 48/sin 22 = 5.95.

2009-08-13 07:58:54 補充：
Cannot answer so many at one time, sorry.

2009-08-13 14:17:07 補充：
Will finish them before Sunday.

2009-08-13 19:25:07 補充：
Glad to see that someone has completed all questions, so will not pursue further.