Chapter 8, Problem 6CT

To determine

The solution of a triangle $A={55}^0,C={20}^0,a=4$ .

Answer to Problem 6CT

Solution:

The other two sides of triangle are $b=4.7167,c=1.6701$, and the third angle is $B={105}^0$ .

Explanation of Solution

Given information:

The two angles and one side of triangle are $A={55}^0,C={20}^0$, and $a=4$ .

To solve the triangle with twoangles and one side given, use the sum of the angles of the triangle.

Here, $A={55}^0$ and $C={20}^0$ .

To find the angle $B$ :

  $A+B+C={180}^0$

  $\Rightarrow {55}^0+B+{20}^0={180}^0$

  $\Rightarrow {75}^0+B={180}^0$

  $\Rightarrow B={180}^0-{75}^0$

  $\Rightarrow B={105}^0$

Now, by law of sine,

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

To find the side $b$ by using the law of sine rule:

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

  $\Rightarrow \frac{\sin {55}^0}{4}=\frac{\sin {105}^0}{b}$

  $\Rightarrow b=\frac{4\cdot \sin {105}^0}{\sin {55}^0}$

  $\Rightarrow b=4.7167$

Therefore, the secondside $b=4.7167$ .

To find the side $c$ by using the law of sine rule:

  $\frac{\sin A}{a}=\frac{\sin C}{c}$

  $\Rightarrow \frac{\sin {55}^0}{4}=\frac{\sin {20}^0}{c}$

  $\Rightarrow c=\frac{4\cdot \sin {20}^0}{\sin {55}^0}$

  $\Rightarrow c=1.6701$

Therefore, the third side $c=1.6701$ .

Thus, the other two sides of triangle are $b=4.7167,c=1.6701$, andthe third angle is $B={105}^0$ .