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.
Found 4 solutions by mananth, ikleyn, CPhill, n2:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
.


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

​


Answer by ikleyn(53560) About Me  (Show Source):
You can put this solution on YOUR website!
.
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~


        @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%28downhill%29 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%28uphill%29 from B the foot of the aerial".

        I came to bring a correct solution. 


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%28232.0790446%29 =  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%28102%5Eo%29%2A%28AB%2FAC%29 = 0.97814760073%2A%289.6%2F15.234%29 = 0.616398646.


Hence,  ∠ACB = arcsin(0.616398646) = 38°  (rounded).

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



Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
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

Answer by n2(36) About Me  (Show Source):
You can put this solution on YOUR website!
.
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~


@CPhill copy-pasted the solution by @mananth and placed it under his name (even without acknowledgment).

So, both "solutions" by @CPhill and by @mananth are identical and both incorrectly interpret the given conditions,
so both solutions are incorrect. The problem says
"A stay connects the top of the aerial A to a point C on the ground 10m highlight%28downhill%29 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%28uphill%29 from B the foot of the aerial".

For correct solution, see the post by @ikleyn at this spot.

Ignore both posts by @CPhill and @mananth - their solutions both are irrelevant to the given problem.