Chapter 3.4, Problem 24P

To determine

To Find: the measure of each angle of the given triangle.

Answer to Problem 24P

  ${92}^{\circ },{46}^{\circ }\text{and }{42}^{\circ }$

Explanation of Solution

Given:

In a triangle ABC, angle A is twice large as B and angle B is $4^{\circ }$ larger than angle C.

  $\langle A=2\langle B\text{and}\langle B-\langle C=4^{\circ }$

Calculation:

Step 1.

Draw diagram as shown below.

  

Step 2.

Since, it is given as.

  $\langle B-\langle C=4^{\circ }$

  $\langle C=\langle B-4^{\circ }$ …… (1)

And,

  $\langle A=2\langle B$ …… (2)

Step 3.

Since, sum of angle of a triangle is ${180}^{\circ }$ therefore.

  $\langle A+\langle B+\langle C={180}^{\circ }$

Substitute, $\langle A=2\langle B$ and $\langle C=\langle B-4^{\circ }$ in above equation.

  $\begin{array}{l}\langle A+\langle B+\langle C={180}^{\circ }\\ 2\langle B+\langle B+\langle B-4={180}^{\circ }\\ 4\langle B-4^{\circ }=180\\ 4\langle B=184\\ \langle B={46}^{\circ }\end{array}$

Step 4.

Substitute, $\langle B={46}^{\circ }$ in equation (2) .

  $\begin{array}{l}\langle A=2\langle B\\ \langle A=2\times 46\\ \langle A={92}^{\circ }\end{array}$

Step 5.

Substitute $\langle B={46}^{\circ }$ in equation (1).

  $\begin{array}{l}\langle C=\langle B-4^{\circ }\\ \langle C=46-4\\ \langle C={42}^{\circ }\end{array}$

Therefore angles of the triangle are ${92}^{\circ },{46}^{\circ }\text{and }{42}^{\circ }$

So the answer is ${92}^{\circ },{46}^{\circ }\text{and }{42}^{\circ }$