Chapter 8, Problem 4CT

To determine

To calculate: Thethree remaining parts of the trianglebelow

Answer to Problem 4CT

Solution:

The two sides of the triangle are $b=6.8519,c=16.2974$, and the angle $C={117}^0$.

Explanation of Solution

Given information:

The triangle

Formula used:

The sum of the angles of any triangle equals $180°$, that is, $\left(A+B+C=180°\right)$.

Law of sine:

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

Calculation:

To solve the triangle when one side and two angles are given, use the law of Sines.

To find the third angle, use the formula $\left(A+B+C={180}^0\right)$.

Here, $A={41}^0,B={22}^0$ and $a=12$.

Substituting the values of $A={41}^0$ and $B={22}^0$.

$\Rightarrow {41}^0+{22}^0+C={180}^0$.

$\Rightarrow {63}^0+C={180}^0$

$\Rightarrow C={117}^0$

Therefore, the third angle $C={117}^0$.

Find the other two sides of the triangle by using the law of Sines.

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

$\Rightarrow \frac{\sin {41}^0}{12}=\frac{\sin {22}^0}{b}$

$\Rightarrow b=\frac{12\cdot \sin {22}^0}{\sin {41}^0}$

$\Rightarrow b=\frac{12\cdot 0.3746}{0.65606}$

$\Rightarrow b=6.8519$

Therefore, the side $b=6.8519$.

To find the side $c$:

Again, by using the law of sine rule,

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

$\Rightarrow \frac{\sin {41}^0}{12}=\frac{\sin {117}^0}{c}$

$\Rightarrow c=\frac{12\cdot \sin {117}^0}{\sin {41}^0}$

$\Rightarrow c=\frac{12\cdot \left(0.8910\right)}{\left(0.65606\right)}$

$\Rightarrow c=\frac{10.69208}{\left(0.65606\right)}$

$\Rightarrow c=16.2974$

Therefore, the side $c=16.2974$.

Therefore, the two sides are $b=6.8519,c=16.2974$, and $C={117}^0$.