In Fig. 6.15, $ \angle \mathrm{PQR}=\angle \mathrm{PRQ} $, then prove that $ \angle \mathrm{PQS}=\angle \mathrm{PRT} $
"
Given:
$\angle PQR=\angle PRQ$.
To do:
We have to prove that $\angle PQS=\angle PRT$.
Solution:
$SQRT$ is a line.
We know that,
The sum of the measures of the angles in linear pairs is always $180^o$.
$\angle PQS+\angle PQR=180^o$ (as they are linear pairs)
$\angle PRT+\angle PRQ=180^o$ (as they are linear pairs)
Therefore,
$\angle PQR=180^o-\angle PQS$..(i)
$\angle PRQ=180^o-\angle PRT$....(ii)
Since,
$\angle PQR=\angle PRQ$
We get, by equating
$180^o-\angle PQS=180^o-\angle PRT$
This implies,
$\angle PQS=\angle PRT$.
- Related Articles
- In the given figure, $\angle PQR=\angle PRT$. Prove that $\angle PQS=\angle PRT$."\n
- In Fig 7.51, PR \( > \) PQ and \( \mathrm{PS} \) bisects \( \angle \mathrm{QPR} \). Prove that \( \angle \mathrm{PSR}>\angle \mathrm{PSQ} \)."\n
- In Fig. 6.44, the side \( \mathrm{QR} \) of \( \triangle \mathrm{PQR} \) is produced to a point \( \mathrm{S} \). If the bisectors of \( \angle \mathrm{PQR} \) and \( \angle \) PRS meet at point \( T \), then prove that \( \angle \mathrm{QTR}=\frac{1}{2} \angle \mathrm{QPR} \)"\n
- In Fig. 6.42, if lines \( \mathrm{PQ} \) and \( \mathrm{RS} \) intersect at point \( \mathrm{T} \), such that \( \angle \mathrm{PRT}=40^{\circ}, \angle \mathrm{RPT}=95^{\circ} \) and \( \angle \mathrm{TSQ}=75^{\circ} \), find \( \angle \mathrm{SQT} \)."\n
- In Fig. 6.39, sides \( \mathrm{QP} \) and \( \mathrm{RQ} \) of \( \triangle \mathrm{PQR} \) are produced to points \( \mathrm{S} \) and T respectively. If \( \angle \mathrm{SPR}=135^{\circ} \) and \( \angle \mathrm{PQT}=110^{\circ} \), find \( \angle \mathrm{PRQ} \)."\n
- In Fig. 6.30, if \( \mathrm{AB} \| \mathrm{CD} \), EF \( \perp \mathrm{CD} \) and \( \angle \mathrm{GED}=126^{\circ} \), find \( \angle \mathrm{AGE}, \angle \mathrm{GEF} \) and \( \angle \mathrm{FGE} \)."\n
- In Fig. 7.21, \( \mathrm{AC}=\mathrm{AE}, \mathrm{AB}=\mathrm{AD} \) and \( \angle \mathrm{BAD}=\angle \mathrm{EAC} \). Show that \( \mathrm{BC}=\mathrm{DE} \)."\n
- In figure below, if \( \angle 1=\angle 2 \) and \( \triangle \mathrm{NSQ} \cong \triangle \mathrm{MTR} \), then prove that \( \triangle \mathrm{PTS} \sim \triangle \mathrm{PRQ} . \)"
- In Fig. 6.41, if \( \mathrm{AB} \| \mathrm{DE}, \angle \mathrm{BAC}=35^{\circ} \) and \( \angle \mathrm{CDE}=53^{\circ} \), find \( \angle \mathrm{DCE} \)."\n
- In \( \triangle \mathrm{PQR}, \quad \angle \mathrm{P}=\angle \mathrm{Q}+\angle \mathrm{R}, \mathrm{PQ}=7 \) and \( \mathrm{QR}=25 \). Find the perimeter of \( \triangle \mathrm{PQR} \).
- In Fig. 6.13, lines \( \mathrm{AB} \) and \( \mathrm{CD} \) intersect at \( \mathrm{O} \). If \( \angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ} \) and \( \angle \mathrm{BOD}=40^{\circ} \), find \( \angle \mathrm{BOE} \) and reflex \( \angle \mathrm{COE} \)."\n
- In Fig. \( 6.40, \angle \mathrm{X}=62^{\circ}, \angle \mathrm{XYZ}=54^{\circ} \). If \( \mathrm{YO} \) and \( Z \mathrm{O} \) are the bisectors of \( \angle \mathrm{XYZ} \) and \( \angle \mathrm{XZY} \) respectively of \( \triangle \mathrm{XYZ} \) find \( \angle \mathrm{OZY} \) and \( \angle \mathrm{YOZ} \)."\n
- In a parallelogram \( \mathrm{ABCD}, \angle \mathrm{A}: \angle \mathrm{B}=2: 3, \) then find angle \( \mathrm{D} \).
- \( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \). If \( 2 \angle \mathrm{P}=3 \angle \mathrm{Q} \) and \( \angle C=100^{\circ} \), find \( \angle B \).
- Choose the correct answer from the given four options:If in triangles \( \mathrm{ABC} \) and \( \mathrm{DEF}, \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}} \), then they will be similar, when(A) \( \angle \mathrm{B}=\angle \mathrm{E} \)(B) \( \angle \mathrm{A}=\angle \mathrm{D} \)(C) \( \angle \mathrm{B}=\angle \mathrm{D} \)(D) \( \angle \mathrm{A}=\angle \mathrm{F} \)
Kickstart Your Career
Get certified by completing the course
Get Started