Chapter 9, Problem 4CR

a

To determine

To find:The value of x

Answer to Problem 4CR

The value of $x={20}^0$

Explanation of Solution

Given information:The following angles are given

  $\angle A={40}^0$

  $\angle B={(3x+10)}^0$

  $ext\angle C={(7x-30)}^0$

Formula used:The sum of two angles of a triangle is equal to the value f the exterior of the third angle

Calculation:We know that in triangle ABC,

  $\angle A+\angle B=ext\angle C$

Thus, substituting the values of these angles, we get

  ${40}^0+{(3x+10)}^0={(7x-30)}^0\to x={20}^0$

Hence, we get

  $x={20}^0$

b

To determine

To find:Whether or not triangle ABC is isosceles

Answer to Problem 4CR

Triangle ABC is NOT isosceles

Explanation of Solution

Given information:The following angles are given

  $\angle A={40}^0$

  $\angle B={(3(20)+10)}^0={70}^0$

  $ext\angle C={(7(20)-30)}^0={110}^0$

Formula used:The sum of two angles of a triangle is equal to the value f the exterior of the third angle

We know that a triangle can be isosceles only if any two of the angles are congruent.

Clearly, in triangle ABC, no angles are congruent. Hence, it is not isosceles.