In the given figure, $PR>PQ$ and PS bisects $\angle QPR$. Prove that $ \angle P S R>\angle P S Q $.
"
Given:
$PR>PQ$ and PS bisects $\angle QPR$.
To do:
We have to prove that \( \angle P S R>\angle P S Q \).
Solution:
We know that,
Angle opposite to the larger side is greater angle.
This implies,
\( \angle P Q R>\angle P R Q \)
\( \angle Q P S=\angle R P S \) (PS bisects $\angle QPR$)
Let \( \angle Q P S=\angle R P S=x \)
In $triangle PQS$,
$\angle PSR=\angle PQR+x$......(i) (Exterior angle property)
In $triangle PSR$,
$\angle PSQ=\angle PRQ+x$......(ii) (Exterior angle property)
\( \angle P Q R>\angle P R Q \) (Given)
Adding $x$ on both sides,
\( \angle P Q R + x >\angle P R Q + x \)
\( \angle P S R>\angle P S Q \)
Hence proved.
- Related Articles
- In Fig 7.51, PR \( > \) PQ and \( \mathrm{PS} \) bisects \( \angle \mathrm{QPR} \). Prove that \( \angle \mathrm{PSR}>\angle \mathrm{PSQ} \)."\n
- In the adjoining figure, $P R=S Q$ and $S R=P Q$.a) Prove that $\angle P=\angle S$.b) $\Delta SOQ \cong \Delta POR$."\n
- In the given figure, $PS$ is the bisector of $\angle QPR$ of $∆PQR$. Prove that $\frac{QS}{SR}=\frac{PQ}{PR}$"
- $PQRS$ is a trapezium in which $PQ||SR$ and $\angle P=130^{\circ}, \angle Q=110^{\circ}$. Then find $\angle R$ and $\angle S$.
- In the figure, tangents \( P Q \) and \( P R \) are drawn from an external point \( P \) to a circle with centre $O$, such that \( \angle R P Q=30^{\circ} . \) A chord \( R S \) is drawn parallel to the tangent \( P Q \). Find \( \angle R Q S \)."\n
- In the figure, common tangents \( P Q \) and \( R S \) to two circles intersect at \( A \). Prove that \( P Q=R S \)."\n
- In quadrilateral $PQRS$, $PQ = PS$ and $PR$ bisccts $\angle P$. Show that $\triangle P R Q \cong \triangle PRS$.What can you say about $QR$ and $SR$?"\n
- In the figure, \( P Q \) is tangent at a point \( R \) of the circle with centre \( O \). If \( \angle T R Q=30^{\circ} \), find \( m \angle P R S \)."\n
- What is the measure of $\angle P$ in the given figure, when $\angle d = 2\angle P$?"\n
- In the adjoining figure, $SP$ and $RQ$ are perpendiculars on the same line $PQ$. Prove that $\Delta P Q S \cong \Delta Q P R$."\n
- In the figure, $AE$ bisects $\angle CAD$ and $\angle B = \angle C$. Prove that $AE \parallel BC$."\n
- In the given figure, $\angle PQR=\angle PRT$. Prove that $\angle PQS=\angle PRT$."\n
- In a triangle \( P Q R, N \) is a point on \( P R \) such that \( Q N \perp P R \). If \( P N \). \( N R=Q^{2} \), prove that \( \angle \mathrm{PQR}=90^{\circ} \).
- In the figure, $PQ \parallel AB$ and $PR \parallel BC$. If $\angle QPR = 102^o$. Determine $\angle ABC$. Give reasons."\n
- In \( \Delta P Q R \), right-angled at \( Q, P Q=3 \mathrm{~cm} \) and \( P R=6 \mathrm{~cm} \). Determine \( \angle P \) and \( \angle R \).
Kickstart Your Career
Get certified by completing the course
Get Started