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 April 27th, 2015, 05:00 PM #1 Newbie   Joined: Feb 2015 From: canada Posts: 10 Thanks: 0 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.
 April 27th, 2015, 05:37 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,756 Thanks: 1407 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 ... $\displaystyle \theta = \arccos\left[\frac{1.1\tan(51^\circ)}{2}\right] \approx 47^\circ$
 April 27th, 2015, 05:53 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,822 Thanks: 1048 Math Focus: Elementary mathematics and beyond 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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# brit and tara are standing 8.8m apart

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