Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 222: 75

Answer

$$\theta\approx18.4^\circ$$

Work Step by Step

$$(l_1):x+y=9\hspace{2cm}(l_2):2x+y=-1$$ $$(l_1):y=-x+9\hspace{2cm}(l_2):y=-2x-1$$ Let's call the angle between the pair of lines $\theta$. We need to apply the formula built in Exercise 74 $$\tan\theta=\frac{m_2-m_1}{1+m_1m_2}$$ in which $m_1$ is the slope of $l_1$ and $m_2$ is the slope of $l_2$. However, since $\tan\theta\gt0$, so $$\tan\theta=\Big|\frac{m_2-m_1}{1+m_1m_2}\Big|$$ The slope of line $(l_1)$: $m_1=-1$ The slope of line $(l_2)$: $m_2=-2$ Therefore, $$\tan\theta=\Big|\frac{-2-(-1)}{1+(-2)(-1)}\Big|$$ $$\tan\theta=\Big|-\frac{1}{3}\Big|$$ $$\tan\theta=\frac{1}{3}$$ So, $$\theta\approx18.4^\circ$$
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