Chapter 12.5, Problem 2CYU

To determine

To find: To round side lengths to the nearest tenth and angles to the nearest degree.

Answer to Problem 2CYU

The angle of triangle ABC.

In triangle ABC

Angle $\angle BAC$ = $\angle A$ = ${112}^{\circ }$

And Angle $\angle BCA$ = $\angle C$ = ${27}^{\circ }$

Angle $\angle ABC=\angle B$ = ${41}^{\circ }$

Explanation of Solution

Given information:

Sides of the triangle ABC

   a = 20

   b= 14

   c = 10

Calculation:

To find the length of $\angle ABC$ = $\angle B$ , we use the cosine formula for any triangle:

  $b^2=a^2+c^2-2ac\cos B$

  c = 10

  b= 14

  a = 3

So

  $\begin{array}{l}{14}^2={20}^2+{10}^2-2\times 20\times 10\cos B\\ \Rightarrow 196=400+100-400\mathrm{Cos}B\\ \Rightarrow 196=500-400\mathrm{Cos}B\\ \Rightarrow \mathrm{Cos}B=\frac{500-196}{400}\\ \Rightarrow \mathrm{Cos}B=\frac{304}{400}\\ \Rightarrow \mathrm{Cos}B=\frac{304}{400}\\ \Rightarrow \mathrm{Cos}B=0.76\\ \Rightarrow B={\cos }^{-1}(0.76)\\ \Rightarrow B=\text{40}\text{.53580211}\\ \Rightarrow B={\text{41}}^{\circ }\end{array}$

Similarly

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ {20}^2={14}^2+{10}^2-2\times 14\times 10\cos A\\ \Rightarrow 400=196+100-280\mathrm{Cos}A\\ \Rightarrow 400=296-280\mathrm{Cos}A\\ \Rightarrow \mathrm{Cos}A=\frac{296-400}{280}\\ \Rightarrow \mathrm{Cos}A=\frac{-104}{280}\\ \Rightarrow \mathrm{Cos}A=-0.3714\\ \Rightarrow A={\cos }^{-1}(-0.3714)\\ \Rightarrow A=\text{111}\text{.80198484}\\ \Rightarrow A={\text{112}}^{\circ }\\ \end{array}$

As we know

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

So

  $\begin{array}{l}A+B+C={180}^{\circ }\\ \Rightarrow C={180}^{\circ }-(A+B)\\ \Rightarrow C={180}^{\circ }-({112}^{\circ }+{41}^{\circ })\\ \Rightarrow C={180}^{\circ }-{153}^{\circ }\\ \Rightarrow C={27}^{\circ }\\ \end{array}$

In triangle ABC

Angle $\angle BAC$ = $\angle A$ = ${112}^{\circ }$

And Angle $\angle BCA$ = $\angle C$ = ${27}^{\circ }$

Angle $\angle ABC=\angle B$ = ${41}^{\circ }$