Chapter 15.3, Problem 7PSA

(a)

To determine

To Find: The largest and the smallest angle

Answer to Problem 7PSA

  $\begin{array}{l}\text{⇒}\angle Y\text{ is largest angle }\\ \text{⇒}\angle X\text{ is the smallest angle }\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\to \Delta WXY\\ \to WX=10\\ \to WY=4\\ \to XY=7\end{array}$

Formula Used:

Pythagorean Theorem:

  $\begin{array}{l}\text{ for acute }\Delta \\ a^2+b^2>c^2\end{array}$

  $\begin{array}{l}\text{for obtuse triangle}\\ a^2+b^2<c^2\end{array}$

  $\begin{array}{l}\text{For right angle triangles}\\ {\text{a}}^2+b^2=c^2\end{array}$

Calculation:

  

  $\begin{array}{l}\to WX=10\\ \to WY=4\\ \to XY=7\\ \text{⇒}\angle Y\text{ is largest angle }\left(\text{Opposite }\angle \text{ of the larger side is the larger }\angle \right)\\ \text{⇒}\angle X\text{ is the smallest angle }\left(\text{Opposite }\angle \text{ of the smaller side is the smaller }\angle \right)\end{array}$

Conclusion:

Thus the $\angle Y\text{ is largest angle ,}\angle X\text{ is the smallest angle }$

(b)

To determine

To Find: Type of the triangle

Answer to Problem 7PSA

  $\text{Obtuse }\Delta$

Explanation of Solution

Given:

  $\begin{array}{l}\to \Delta WXY\\ \to WX=10\\ \to WY=4\\ \to XY=7\end{array}$

Formula Used:

Pythagorean Theorem:

  $\begin{array}{l}\text{ for acute }\Delta \\ a^2+b^2>c^2\end{array}$

  $\begin{array}{l}\text{for obtuse triangle}\\ a^2+b^2<c^2\end{array}$

  $\begin{array}{l}\text{For right angle triangles}\\ {\text{a}}^2+b^2=c^2\end{array}$

Calculation:

  

  $\begin{array}{l}\text{⇒}W{Y}^2+X{Y}^2=W{X}^2\left(\text{pythagoras theorem}\right)\\ \text{⇒}{a}^2+b^2<c^2\left(\text{Obtuse }\Delta \right)\\ \text{⇒}{4}^2+7^2<{10}^2\left(\text{Using triangle values}\right)\\ \text{⇒}65<100\\ \text{⇒}\Delta WXY\text{ is obtuse triangle}\end{array}$

Conclusion:

Thus the $\Delta WXY$ is $\text{Obtuse }\Delta$