Question 1201106
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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.
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        @mananth incorrectly interpreted the given conditions,

        so his solution is incorrect. The problem says 

        "A stay connects the top of the aerial A to a point C on the ground 10m {{{highlight(downhill)}}} from B the foot of the aerial",

        but @mananth interpreted in his solution as if

        "A stay connects the top of the aerial A to a point C on the ground 10m {{{highlight(uphill)}}} from B the foot of the aerial".


        I came to bring a correct solution.



<pre>
We have triangle ABC with side AB of 9.6 m, side BC of 10 m
and angle ABC of 90° + 12° = 102°.

We want to find side AC opposite to angle ABC.
 

Use the law off cosines

AC^2  = AB^2 + BC^2 −2*AB*BC*cos(∠ABC) = 

      = 9.6^2 + 10^2 - 2*10*9.6*cos(102°) = 92.16 + 100 - 2*9.6*10*(-0.20791169081) = 232.0790446.

So, length of the stay AC  is  {{{sqrt(232.0790446)}}} =  15.234 m.


To find the angle ACB, which the stay makes with the ground, use the Law of sines 

    sin(∠ACB)/AB = sin(∠ABC)/AC,

    sin(∠ACB) = {{{sin(102^o)*(AB/AC)}}} = {{{0.97814760073*(9.6/15.234)}}} = 0.616398646.


Hence,  ∠ACB = arcsin(0.616398646) = 38°  (rounded).
</pre>

At this point, the problem is solved correctly and completely.