Chapter 15, Problem 3RP

(a)

To determine

To Find: The greater angle in the given figure.

Answer to Problem 3RP

  $\angle CBD$

Explanation of Solution

Given:

  $\begin{array}{l}AB=10\\ BC=10\\ CD=6\\ AD=5\end{array}$

  

Formula Used:

In a triangle side opposite to greater angle is always greater and opposite to smallest angel is always smallest or vice-versa.

Angle-Sum Property: In a triangle the sum of all interior angle is equal to 180.

Calculation:

  $\begin{array}{l}In\Delta ABC\\ AB=10\\ BC=10\\ CD=6\\ AD=5\\ AC=5+6=11\\ AD<CD\\ \because \overline{AD}\text{ is opposite to }\angle ABD\\ and\overline{CD}\text{ is opposite to }\angle CBD\\ \therefore \angle ABD<\angle CBD\end{array}$

Conclusion:

Thus the required angle is $\angle CBD$

(b)

To determine

To Find: The greater angle in the given figure.

Answer to Problem 3RP

  $\angle X$

Explanation of Solution

Given:

  $\begin{array}{l}YZ=7\\ XZ=6\\ XY=5\end{array}$

  

Formula Used:

In a triangle side opposite to greater angle is always greater and opposite to smallest angel is always smallest or vice-versa.

Angle-Sum Property: In a triangle the sum of all interior angle is equal to 180.

Calculation:

  $\begin{array}{l}In\Delta XYZ\\ YZ=7\\ XZ=6\\ XY=5\\ XYorXZ<YZ\\ \because \overline{YZ}\text{ is opposite to }\angle X\\ \therefore \angle Yor\angle Z<\angle X\end{array}$

Conclusion:

Thus the required angle is $\angle X$

(c)

To determine

To Find: The greater angle in the given figure.

Answer to Problem 3RP

  $\angle 1$

Explanation of Solution

Given:

  $\begin{array}{l}\angle PQS=\angle 2\\ \angle QSR=\angle 1\end{array}$

  

Formula Used:

In a triangle side opposite to greater angle is always greater and opposite to smallest angel is always smallest or vice-versa.

Exterior-Angle Property: In a triangle sum of two interior angle is equal to opposite exterior angle.

Calculation:

  $\begin{array}{l}In\Delta PQR\&\Delta PQS\\ \angle PQS=\angle 2\\ \angle QSR=\angle 1\\ \angle QPS+\angle PQS=\angle QSR\\ \Rightarrow \angle P+\angle 2=\angle 1\\ \therefore \angle 1>\angle Por\angle 2\end{array}$

Conclusion:

Thus the required angle is $\angle 1$