Question #277258

A vertical aerial AB 12.65m high stands on ground which is inclined 18°32 15to the horizontal. A stay connects the top of the aerial A to a point Con the ground 20.0 m uphill from Bthe foot of the aerial.

Determine:

the length of the stay .

the angle the stay makes with the ground


1
Expert's answer
2021-12-09T04:46:03-0500


Given AB=12.65m,BC=20.0m,α=18°3215.AB=12.65m, BC=20.0m, \alpha=18\degree32'15''.


β=90°α\beta=90\degree-\alpha

Then


ABC=β=90°α\angle ABC=\beta=90\degree-\alpha

Triangle ACB


AB=12.65,BC=20,AB=12.65, BC=20, ''

ABC=90°α=90°18°3215\angle ABC=90\degree-\alpha=90\degree-18\degree32'15''

a) Law of Cosines


AC2=AB2+BC22(AB)(BC)cosABC{AC}^2={AB}^2+{BC}^2-2(AB)(BC)\cos\angle ABC

AC=12.652+2022(12.65)(20)sin(18°3215)AC=\sqrt{{12.65}^2+{20}^2-2(12.65)(20)\sin(18\degree32'15'')}

AC=19.98mAC=19.98m

2) Law of Sines


sinACBAB=sinABCAC\dfrac{\sin\angle ACB}{AB}=\dfrac{\sin \angle ABC}{AC}

ACB=sin1(ABsinABCAC)\angle ACB=\sin^{-1}(\dfrac{AB\sin \angle ABC}{AC})

ACB=sin1(12.65cos(18°3215)19.98)\angle ACB=\sin^{-1}(\dfrac{12.65\cos (18\degree32'15'')}{19.98})

ACB=36.89°\angle ACB=36.89\degree

ACB=36°5325\angle ACB=36\degree53'25''

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