Chapter 13.5, Problem 3CYU

To determine

To find: To find the angles of the triangle to the nearest degree.

Answer to Problem 3CYU

In triangle ABC

Angle $\angle BAC={18}^{\circ }$

Angle $\angle ACB={29}^{\circ }$

Angle $\angle ABC={133}^{\circ }$

Explanation of Solution

Given information:

The length three sides of triangle ABC.

AB= c = 8

BC= a = 5

CA= b = 12

Calculation:

To find the angle A, B and C we use the cosine formula for any triangle:

  $b^2=a^2+c^2-2ac\cos B$

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow {12}^2=5^2+8^2-2\times 5\times 8\times \cos B\\ \Rightarrow 144=25+64-80\mathrm{Cos}B\\ \Rightarrow 144=89-80\mathrm{Cos}B\\ \Rightarrow \mathrm{Cos}B=\frac{89-144}{80}\\ \Rightarrow \mathrm{Cos}B=\frac{-55}{80}\\ \Rightarrow \mathrm{Cos}B=\frac{-55}{80}\\ \Rightarrow \mathrm{Cos}B=-0.6875\\ \Rightarrow B={\cos }^{-1}(-0.6875)\\ \Rightarrow B={133.43}^{\circ }\\ \Rightarrow B={133}^{\circ }\end{array}$

Also as we know that-

  $\begin{array}{l}{a}^2=b^2+c^2-2ac\cos A\\ \Rightarrow 5^2={12}^2+8^2-2\times 12\times 8\times \cos A\\ \Rightarrow 25=144+64-192\mathrm{Cos}A\\ \Rightarrow 25=208-192\mathrm{Cos}A\\ \Rightarrow \mathrm{Cos}A=\frac{208-25}{192}\\ \Rightarrow \mathrm{Cos}A=\frac{183}{192}\\ \Rightarrow \mathrm{Cos}A=0.9531\\ \Rightarrow A={\cos }^{-1}(0.9531)\\ \Rightarrow A={17.61}^{\circ }\\ \Rightarrow A={18}^{\circ }\end{array}$

As we know

The sum of the angles of any triangle is ${180}^{\circ }$

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow C={180}^{\circ }-(A+B)\\ \Rightarrow C={180}^{\circ }-({18}^{\circ }+{133}^{\circ })\\ \Rightarrow C={180}^{\circ }-{151}^{\circ }\\ \Rightarrow C={29}^{\circ }\end{array}$

In triangle ABC

Angle $\angle BAC={18}^{\circ }$

Angle $\angle ACB={29}^{\circ }$

Angle $\angle ABC={133}^{\circ }$