Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - P.S. Problem Solving - Page 557: 7


a) $A=100 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} $ b) $A=50 \sin \theta$ and $\theta =\dfrac{\pi}{2}$

Work Step by Step

a) Let $\dfrac{\theta}{2}$ be half the angle of the triangle with base $b$ and height $h$. Area, $A=\dfrac{1}{2}\ b \ h...(1)$ We will use the sine and cosine functions to compute base $b$ and height $h$. Now, $\sin \dfrac{\theta}{2}=\dfrac{b/2}{10} \implies b=20 \sin \dfrac{\theta}{2}$ and $\cos \dfrac{\theta}{2}=\dfrac{h}{10} \implies h=10 \cos \dfrac{\theta}{2}$ Equation (1) becomes: $Area, A=\dfrac{1}{2} (20 \sin \dfrac{\theta}{2}) (10 \cos \dfrac{\theta}{2}) =100 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} $ b) $Area, A=50 (2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}) $ $\sin 2x=2 \sin x \cos x$ Now, $A=50 \sin \theta$ To attain a maximum area, $\sin \theta$ must equal $1$. Therefore, $\theta =\dfrac{\pi}{2}$.
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