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: 76

Answer

$$\theta\approx80.8^\circ$$

Work Step by Step

$$(l_1):5x-2y+4=0\hspace{2cm}(l_2):3x+5y=6$$ $$(l_1):2y=5x+4\hspace{2cm}(l_2):5y=-3x+6$$ $$(l_1):y=\frac{5}{2}x+2\hspace{2cm}(l_2):y=-\frac{3}{5}x+\frac{6}{5}$$ 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 $(l_1)$: $m_1=\frac{5}{2}$ The slope of line $(l_2)$: $m_2=-\frac{3}{5}$ Therefore, $$\tan\theta=\Big|\frac{-\frac{3}{5}-\frac{5}{2}}{1+(-\frac{3}{5})(\frac{5}{2})}\Big|$$ $$\tan\theta=\Big|\frac{-\frac{31}{10}}{1-\frac{15}{10}}\Big|$$ $$\tan\theta=\Big|\frac{-\frac{31}{10}}{-\frac{5}{10}}\Big|$$ $$\tan\theta=\Big|\frac{31}{5}\Big|=\frac{31}{5}$$ So, $$\theta\approx80.8^\circ$$
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