Chapter 6, Problem 29CR

To determine

To identify:

For solving given triangle law of cosine is used or not

Answer to Problem 29CR

Angles sof $△ABC$ be $\angle A=77.52°,\angle B=38.48°,a=14.12$

Explanation of Solution

Given information:

  $b=9,c=13,C=64°$

Calculation:

Let’s consider,

  $b=9,c=13,C=64°$

For solving the triangle by using law of sine, put the values,

  $\begin{array}{l}\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{9}{\sin B}=\frac{13}{\sin 64°}\\ \sin B=\frac{9}{13}\left(\sin 64°\right)\\ \sin B=\frac{9}{13}\left(0.89987\right)\\ \sin B=\frac{8.089}{13}\\ \sin B=0.622\\ B={\sin }^{-1}\left(0.622\right)\\ B=38.476°\\ B=38.48°\end{array}$

Sum of three angle is, $△ABC$

  $\begin{array}{l}\angle A+\angle B+\angle C=180°\\ \angle A+38.48°+64°=180°\\ \angle A=180°-102.48°\\ \angle A=77.52°\end{array}$

Ny using law of cosine,

  $\begin{array}{l}\cos A=\frac{b^2+c^2-c^2}{2bc}\\ a^2=b^2+c^2-2bc\cos A\end{array}$

Put the values in above equation,

  $\begin{array}{l}{a}^2={\left(9\right)}^2+{\left(13\right)}^2-2\left(9\right)\left(13\right)\cos \left(77.52°\right)\\ a^2=81+169-234\left(0.2160\right)\\ a^2=250-50.54\\ a^2=199.4\\ a=\sqrt{199.46}\\ a=14.12\end{array}$

Therefore, value of $\angle A=77.52°,\angle B=38.48°,a=14.12$