Trigonometry (11th Edition) Clone

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

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

Answer

$d = 3.454~m$

Work Step by Step

We can convert angle $\alpha$ to degrees: $\alpha = 37'48''$ $\alpha = (\frac{37}{60}+\frac{48}{3600})^{\circ}$ $\alpha = 0.63^{\circ}$ We can convert angle $\beta$ to degrees: $\beta = 42'03''$ $\beta = (\frac{42}{60}+\frac{3}{3600})^{\circ}$ $\beta = 0.7008^{\circ}$ It is given that $b = 2.000~cm$ We can use the expression for the distance $d$ in part (a) to find the distance $d$ from point P to point Q: $d = \frac{b}{2~tan(\frac{\alpha}{2})}+\frac{b}{2~tan(\frac{\beta}{2})}$ $d = \frac{2.000~cm}{2~tan(\frac{0.63^{\circ}}{2})}+\frac{2.000~cm}{2~tan(\frac{0.7008^{\circ}}{2})}$ $d = 181.8895~cm+163.5133~cm$ $d = 3.454~m$
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