Trigonometry (10th Edition)

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises: 20

Answer

To solve the triangle here is to find the angle $B$, side $a$ and side $b$. $$B=67.75^\circ, a=22.04mm, b=37.50mm$$

Work Step by Step

$$A=32^\circ57′,c=39.81mm,C=79^\circ18′ $$ $$B=?, a=?, b=?$$ 1) Change the angle to complete degrees: Remember that $60′=1^\circ$ Therefore, $$A=32^\circ57′=32^\circ+(\frac{57}{60})^\circ=32^\circ+0.95^\circ=32.95^\circ$$ $$C=79^\circ18′=79^\circ+(\frac{18}{60})^\circ=79^\circ+0.3^\circ=79.3^\circ$$ 2) Find $B$: The sum of 3 angles in any triangle is $180^\circ$. That means, $$32.95^\circ+B+79.3^\circ=180^\circ$$ $$112.25^\circ+B=180^\circ$$ $$B=67.75^\circ$$ 3) Find $a$ and $c$ The law of sines for a triangle with side $a,b,c$ is $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$ First, to find $a$, we notice that the values of $c, C$ and $A$ are already known. So $a$ can be found using this $$\frac{a}{\sin A}=\frac{c}{\sin C}$$ $$a=\frac{c\sin A}{\sin C}$$ $$a=\frac{39.81\sin32.95^\circ}{\sin79.3^\circ}$$ $$a=\frac{39.81\times0.544}{0.983}$$ $$a\approx22.04mm$$ To find $b$, we continue to notice that the values of $c, C$ and $B$ are already known. So $b$ can be found using this $$b\sin B=c\sin C$$ $$b=\frac{c\sin B}{\sin C}$$ $$b=\frac{39.81\sin67.75^\circ}{\sin79.3^\circ}$$ $$b=\frac{39.81\times0.926}{0.983}$$ $$b\approx37.50mm$$ In conclusion: $$B=67.75^\circ, a=22.04mm, b=37.50mm$$
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