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 Britâ€™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: 2,947 Thanks: 1552 
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: 7,943 Thanks: 1134 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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