SOLUTION: A vertical aerial AB 9.6m high stands on ground which is inclined 12 degrees to the horizontal. A stay connects the top of the aerial A to a point C on the ground 10m downhill from

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Question 1201106: A vertical aerial AB 9.6m high stands on ground which is inclined 12 degrees to the horizontal. A stay connects the top of the aerial A to a point C on the ground 10m downhill from B the foot of the aerial. Determine the length of the stay and the angle the stay makes with the ground.
Answer by mananth(16946) About Me  (Show Source):
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AB =9.6 m
%28alpha%29 = 12 deg. %28beta%29+=+%2890-%28alpha%29%29
90-12 = 78
Law of cosines
AC^2 =AB^2+BC^2 −2(AB)(BC)cos∠ABC
= (9.6)^2 + (10)^2 - 2*(10)*(9.6)*cos (78deg)
=152.24 m
length of the stay = 152.24 m
Law of sines
(sin∠ACB/AB ) = (sin∠ABC/AC)
Find the angle

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