Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 303: 39

Answer

The balloon is $0.485mi$ above the ground.

Work Step by Step

Take a look at the image below. We have $a\parallel AB$. Now recall the characteristics of pairs of angles in parallel lines. As pairs of alternate internal angles are equal, $$\angle A=35^\circ\hspace{2cm}\angle B=31^\circ$$ Also, as $a\parallel AB$, $$\angle L +35^\circ+31^\circ=180^\circ$$ $$\angle L+66^\circ=180^\circ$$ $$\angle L=114^\circ$$ $$\angle A=35^\circ\hspace{.75cm}\angle B=31^\circ\hspace{.7cm}\angle L=114^\circ\hspace{.75cm}AB=1.5mi\hspace{.75cm}LH=?$$ 1) Analysis - To find $LH$, we first need to find $LB$. - To find $LB$, we would use law of sines. 2 groups of side and its opposite angles must be known. - The opposite angles of $LB$ and $AB$, $\angle A$ and $\angle L$ respectively, are already known. 2) Apply law of sines to find $LB$. - The opposite angle of $LB$ is $\angle A$, $\sin A=\sin 35^\circ\approx0.57$ - $AB=1.5mi$, its opposite angle is $\angle L$, $\sin L=\sin 114^\circ\approx0.91$ According to law of sines: $$\frac{LB}{\sin A}=\frac{AB}{\sin L}$$ $$LB=\frac{AB\sin A}{\sin L}$$ $$LB=\frac{1.5mi\times0.57}{0.91}$$ $$LB\approx0.94mi$$ 4) Find $LH$ In $\triangle LHB$: $LH \bot HB$ $$LH=LB\sin B$$ $$LH=0.94mi\times\sin 31^\circ$$ $$LH\approx0.485mi$$ Therefore, the balloon is $0.485mi$ above the ground.
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