## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Acute Angles and Right Triangles - Section 2.5 Further Applications of Right Triangles - 2.5 Exercises - Page 89: 41a

#### Answer

$d = \frac{b}{2}~(cot~\frac{\alpha}{2}+~cot~\frac{\beta}{2})$

#### Work Step by Step

Let $d_1$ be the distance from point P to the center of the bar. We can find an expression for $d_1$: $\frac{b/2}{d_1} = tan(\frac{\alpha}{2})$ $d_1 = \frac{b}{2~tan(\frac{\alpha}{2})}$ $d_1 = \frac{b~cot(\frac{\alpha}{2})}{2}$ Let $d_2$ be the distance from point Q to the center of the bar. We can find an expression for $d_2$: $\frac{b/2}{d_2} = tan(\frac{\beta}{2})$ $d_2 = \frac{b}{2~tan(\frac{\beta}{2})}$ $d_2 = \frac{b~cot(\frac{\beta}{2})}{2}$ The total distance $d$ from point P to point Q is $d_1+d_2$. We can find an expression for $d$: $d = d_1+d_2$ $d = \frac{b~cot(\frac{\alpha}{2})}{2}+\frac{b~cot(\frac{\beta}{2})}{2}$ $d = \frac{b}{2}~(cot~\frac{\alpha}{2}+~cot~\frac{\beta}{2})$

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