Question

In the given problem solve the triangle using either the law of sines or law of cosines-

$a=10,b=10,c=15$

Short answer

Required values of the triangle are

$A=138.59°,B=41.29°,C=81.89°$

Step-by-step solution

Step 1. Given information

In the given triangle,

$a=10,b=10,c=15$

We know,

Law of cosines,

$a^2=b^2+c^2-2bc\cos A$ --------$\left(1\right)$

Law of sines,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ ---------$\left(2\right)$

Step 2. Calculation

For $A$, using equation $\left(1\right)$

role="math" localid="1646736804050" $$\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ \Rightarrow {10}^2={10}^2-{15}^2-2\times 10\times 15\cos A \\ \Rightarrow 300\cos A=-225 \\ \Rightarrow \cos A=\frac{-225}{300} \\ \Rightarrow A={\cos }^{-1}\left(-0.75\right) \\ \Rightarrow A=138.59° \end{gathered}$$

For $B$and $C$, using $\left(2\right)$

$$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \Rightarrow \frac{\sin 138.59°}{10}=\frac{\sin B}{10} \\ \Rightarrow \sin B=0.66 \\ \Rightarrow B={\sin }^{-1}\left(0.66\right)=41.29° \end{gathered}$$

Also,

$$\begin{gathered} \frac{\sin B}{b}=\frac{\sin C}{c} \\ \Rightarrow \frac{\sin (41.29°)}{10}=\frac{\sin C}{15} \\ \Rightarrow \sin C=\frac{0.66}{10}\times 15 \\ \Rightarrow \sin C=0.99 \\ \Rightarrow C={\sin }^{-1}\left(0.99\right)=81.89 \end{gathered}$$