Chapter 13.5, Problem 26PPS

a.

To determine

To find: What will be the angles at which the fences of three sides meet.

Answer to Problem 26PPS

  $\begin{array}{l}A={124}^0\\ B={37}^0\\ C={19}^0\end{array}$

Explanation of Solution

Given information:

Let

  $\begin{array}{l}a=1.25\\ b=0.9\\ c=0.5\end{array}$

Using law of Cosines we get

  $a^2=b^2+c^2-2bc\cos A$

On substituting values

We get

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ \Rightarrow {\left(1.25\right)}^2={\left(0.9\right)}^2+{\left(0.5\right)}^2-2\left(0.9\right)\left(0.5\right)\cos A\\ \Rightarrow 1.5625=0.81+0.25-0.9\cos A\\ \Rightarrow 1.5625=1.06-0.9\cos A\\ \Rightarrow 0.9\cos A=1.06-1.5625\\ \Rightarrow 0.9\cos A=-0.5025\\ \Rightarrow \cos A=\frac{-0.5025}{0.9}\\ \Rightarrow \cos A=-0.558\\ \Rightarrow A={\cos }^{-1}\left(-0.558\right)\\ \Rightarrow A={124}^0\end{array}$

Using law of Sines

We get

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {124}^0}{1.25}=\frac{\sin B}{0.9}\\ \Rightarrow \sin B=\frac{0.9\sin {124}^0}{1.25}\\ \Rightarrow \sin B=\frac{0.9\left(0.829\right)}{1.25}\\ \Rightarrow \sin B=\frac{0.7461}{1.25}\\ \Rightarrow \sin B=0.59688\\ \Rightarrow B={\sin }^{-1}\left(0.59688\right)\\ \Rightarrow B={37}^0\end{array}$

We know

Sum of angles of a triangle is ${180}^0$

Using that

We get

  $A+B+C={180}^0$

On substituting values

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {124}^0+{37}^0+C={180}^0\\ \Rightarrow {161}^0+C={180}^0\\ \Rightarrow C={180}^0-{161}^0\\ \Rightarrow C={19}^0\end{array}$

Therefore,

  $\begin{array}{l}A={124}^0\\ B={37}^0\\ C={19}^0\end{array}$