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: 38

Answer

$a\approx26.80$ $;$ $h\approx103.53$ $\angle A=15^{\circ}$

Work Step by Step

The triangle is shown below. Two angles are known. Let $\angle B$ be the right angle and $\angle C$ be the angle marked as $75^{\circ}$. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle A$: $\angle A=180^{\circ}-90^{\circ}-75^{\circ}=15^{\circ}$ Find the sine trigonometric ratio of the angle marked $75^{\circ}$ to obtain $h$: $\sin75^{\circ}=\dfrac{100}{h}$ $h=\dfrac{100}{\sin75^{\circ}}\approx103.53$ Use the Pythagorean Theorem to obtain the unknown cathetus $a$: $a=\sqrt{(103.53)^{2}-100^{2}}\approx26.80$
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