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April 27th, 2015, 05:00 PM   #1
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Question solving triangles

Brit and Tara are standing 8.8 m apart on a dock when they observe a
sailboat moving parallel to the dock. When the boat is equidistant between
both girls, the angle of elevation to the top of its 8.0 m mast is 51 degrees for both observers. Describe how you would calculate the angle, to the nearest degree,between Tara and the boat as viewed from Brit’s position. Justify your reasoning with calculations.

The answer should be 47 degrees, but I can't get the answer.

Last edited by skipjack; April 28th, 2015 at 12:16 AM.
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April 27th, 2015, 05:37 PM   #2
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horizontal distance from either observer to the base of the mast is

$\displaystyle x = \frac{8}{\tan(51^\circ)}$

the base triangle formed by the two observers and the mast bottom, parallel to the water's surface, has sides $x$, $x$, $8.8$ m

let $\theta$ be the measure of each base angle of the isosceles triangle described above.

using the cosine law ...

$\displaystyle \theta = \arccos\left[\frac{1.1\tan(51^\circ)}{2}\right] \approx 47^\circ$
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April 27th, 2015, 05:53 PM   #3
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Did you make a diagram?

To calculate the distance from Tara to the sailboat, $\displaystyle x$, use the tangent ratio:

$\displaystyle \tan51^\circ=\frac8x\Rightarrow x=\frac{8}{\tan51^\circ}$

With $\displaystyle \theta$ being the angle we seek, and using the cosine ratio, we have

$\displaystyle \cos\theta=\frac{4.4}{x}\Rightarrow\theta=\cos^{-1}\left(\frac{4.4}{x}\right)\approx47^\circ$
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