Chapter 15.2, Problem 9PSB

To determine

To Find: The possible value of the required angle.

Answer to Problem 9PSB

  ${70}^0>\angle B>{46}^0$

Explanation of Solution

Given:

  $\angle A={\text{Vertex Angle of isosceles triangle in between 40}}^0{\text{ and 88}}^0$

Formula Used:

Base-Angle Theorem: It states that if two sides of a triangle are congruent then then the angle opposite to the sides are also congruent and vice-versa.

Calculation:

  $\begin{array}{l}\text{Base angle of isosceles traingle are congruent}\text{.}\\ \text{so }\angle \text{B}\\ \frac{\left(180-40\right)}{2}>\angle B>\frac{\left(180-88\right)}{2}\\ \frac{140}{2}>\angle B>\frac{92}{2}\\ {70}^0>\angle B>{46}^0\end{array}$

Conclusion:

Thus, $\angle B{\text{ of isosceles triangle is between 46}}^0{\text{and 70}}^0.$