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Q. 29

Expert-verified
Found in: Page 535

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Solve each triangle.$a=5,b=8,c=9$

The required triangle is

$A=33.90°\phantom{\rule{0ex}{0ex}}B=62.61°\phantom{\rule{0ex}{0ex}}C=83.49°$

See the step by step solution

## Step 1: Given information

$a=5,b=8,c=9$

To find the angles $A,B,C$

## Step 2: Calculation

$ForA:\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(A\right)=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}=\frac{{8}^{2}+{9}^{2}-{5}^{2}}{2×8×9}\phantom{\rule{0ex}{0ex}}=0.83\phantom{\rule{0ex}{0ex}}A={\mathrm{cos}}^{-1}\left(0.83\right)\phantom{\rule{0ex}{0ex}}=33.90°\phantom{\rule{0ex}{0ex}}$

$ForB:\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(B\right)=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{2}+{9}^{2}-{8}^{2}}{2×5×9}\phantom{\rule{0ex}{0ex}}=0.46\phantom{\rule{0ex}{0ex}}B={\mathrm{cos}}^{-1}\left(0.46\right)\phantom{\rule{0ex}{0ex}}=62.61°$

Since we know A and B

$C=180°-A-B\phantom{\rule{0ex}{0ex}}=180°-33.90°-62.61°\phantom{\rule{0ex}{0ex}}=83.49°$