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Q. 28

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.$a=3,b=3,c=2$

The required triangle is

$A=70.73°\phantom{\rule{0ex}{0ex}}B=70.73°\phantom{\rule{0ex}{0ex}}C=38.54°$

See the step by step solution

## Step 1: Given information

$a=3,b=3,c=2$

To find the angles A,B,C

## Step 2: Calculation

$ForA:\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(A\right)=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{2}+{2}^{2}-{3}^{2}}{2×3×2}\phantom{\rule{0ex}{0ex}}=0.33\phantom{\rule{0ex}{0ex}}A={\mathrm{cos}}^{-1}\left(0.33\right)\phantom{\rule{0ex}{0ex}}=70.73°\phantom{\rule{0ex}{0ex}}ForB:\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(B\right)=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\phantom{\rule{0ex}{0ex}}=\frac{{3}^{2}+{2}^{2}-{3}^{2}}{2×3×2}\phantom{\rule{0ex}{0ex}}=0.33\phantom{\rule{0ex}{0ex}}B={\mathrm{cos}}^{-1}\left(0.33\right)\phantom{\rule{0ex}{0ex}}=70.73°\phantom{\rule{0ex}{0ex}}$

Since we know A and B

$C=180°-A-B\phantom{\rule{0ex}{0ex}}=180°-70.73-70.73\phantom{\rule{0ex}{0ex}}=38.54°$