Chapter 6.1, Problem 25E

To determine

To calculate:

Solution of triangle according to law of sines.

Answer to Problem 25E

Triangle with $A=50°,a=10°,b=40°$ characteristics does not exist.

Explanation of Solution

Given information:

  $A=50°,a=10°,b=40°$

Calculation:

A triangle $ABC$ in which $A=50°,a=10,b=40$ ,

  $\begin{array}{l}h=40\left(\sin 50°\right)\\ h=30.64\end{array}$

As $a<b$ it is conclude that two solution exists.

By the law of sines,

  $\begin{array}{l}\frac{b}{\sin B}=\frac{a}{\sin A}\\ SinA=a\left(\frac{\sin B}{b}\right)\\ \sin A=10\left(\frac{\sin B}{40}\right)\\ \sin 45°=\frac{\sin B}{4}\\ 2.82=\sin B\end{array}$

The third angle of triangle,

  $c=180°-A-B$

As value of $B$ does not exist.

Hence, it is concluded that triangle with $A=50°,a=10°,b=40°$ characteristics does not exist.