Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.2 - Trigonometry of Right Triangles - 6.2 Exercises - Page 488: 37


$\angle A=45^{\circ}$ $;$ $\angle B=45^{\circ}$ $;$ $\angle C=90^{\circ}$ $a=16$ $;$ $b=16$ $;$ $h=16\sqrt{2}$

Work Step by Step

The triangle is shown below. Let $h$ be the hypotenuse of the triangle, $a$ be the known cathetus and $b$ the unknown cathetus. Two angles are known. Let $\angle A$ be the known angle and $\angle B$ and $\angle C$ the known ones. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle A$: $\angle A=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$ Use the sine trigonometric ratio of the known angle to find $h$: $\sin45^{\circ}=\dfrac{16}{h}$ $h=\dfrac{16}{\sin45^{\circ}}=\dfrac{16}{\Big(\dfrac{\sqrt{2}}{2}\Big)}=\dfrac{16(2)}{\sqrt{2}}=16\sqrt{2}$ Use the Pythagorean Theorem to obtain the unknown cathetus: $b=\sqrt{(16\sqrt{2})^{2}-16^{2}}=\sqrt{(16^{2})(2)-16^{2}}=\sqrt{512-256}=...$ $...=\sqrt{256}=16$
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