# How do you solve the triangle given m∠B = 73°, a = 7, b = 5?

Feb 18, 2018

Triangle can’t exist as $\sin A > 1$

#### Explanation:

Given : $\hat{B} = {73}^{\circ} , a = 7 , b = 5$

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

$\therefore \sin A = \frac{a \sin B}{b} = \frac{7 \sin 73}{5} = 1.3388$

Since, sine of an angle cannot have more than 1, such a triangle cannot exist.