Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises - Page 73: 9

Answer

$a=571$ m $b=777$ m $B=53^{\circ}40'$

Work Step by Step

Step 1: To find $a$, we use the formula $\sin\theta=\frac{a}{c}$. Step 2: $\sin36^{\circ}20'=\frac{a}{964}$ Step 3: Converting the angle to decimal degrees; $36^{\circ}20'=36\frac{20}{60}^{\circ}=36.3^{\circ}$ Step 4: $a=964\times\sin36.3^{\circ}$ Step 5: Using a calculator, $a\approx570.7$ Step 6: Rounding the answer to three significant degrees, $a\approx571$ m. Step 7: To find $b$, we use the formula $\cos\theta=\frac{b}{c}$. Step 8: $\cos36.3^{\circ}=\frac{b}{964}$ Step 9: $b=964\times\cos36.3^{\circ}$ Step 10: Using a calculator, $b\approx776.9$ Step 11: Rounding the answer to three significant degrees, $b\approx777$ m. Step 12: As $A+B=90^{\circ}$, $B=90^{\circ}-36^{\circ}20'$ Step 13: Solving, $B=53^{\circ}40'$.
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