Chapter 15, Problem 23RP

To determine

To Find: The probability of a triangle being obtuse.

Answer to Problem 23RP

  $\frac{4}{17}$

Explanation of Solution

Given:

  $\Delta ABC$

  $\begin{array}{l}AB=5\\ AC=13\end{array}$

Formula Used:

Triangle-Inequality Theorem: It states that the sum of two sides of a triangle is always greater than the third side.

  $\begin{array}{l}Acute:a^2+b^2>c^2\\ Obtuse:a^2+b^2<c^2\\ Right:a^2+b^2=c^2\end{array}$

  $\text{Probability=}\frac{\text{Total outcomes}}{\text{Total Sample space}}$

Calculation:

  $\begin{array}{l}In\Delta ABC\\ AB=5\\ AC=13\\ BC<AB+AC=13+5=18\\ A{B}^2+A{C}^2=5^2+{13}^2=25+169=194<B{C}^2\\ \text{for being obtuse: }13<BC<18\\ \text{Total sample space for sides }BC=17\\ \text{Total outcomes for being obtuse}=4\\ \text{Probability of being obtuse}=\frac{4}{17}\\ \end{array}$

Conclusion:

Thus the probability of being obtuse is $\frac{4}{17}$ .