In the figure, $PQRS$ is a square and $SRT$ is an equilateral triangle. Prove that $\angle TQR = 15^o$.
"
Given:
$PQRS$ is a square and $SRT$ is an equilateral triangle.
To do:
We have to prove that $PT = QT$.
Solution:
In $\triangle TSP$ and $\triangle TQR$,
$ST = RT$ (Sides of an equilateral triangle)
$SP = PQ$ (Sides of square)
$\angle TSP = \angle TRQ$
Therefore, by SAS axiom,
$\triangle TSP \cong \triangle TQR$
This implies,
$PT = QT$ (CPCT)
In $\triangle TQR$,
$RT = RQ$ (Sides of a square)
$\angle RTQ = \angle RQT$
$\angle TRQ = 60^o + 90^o = 150^o$
$\angle RTQ + \angle RQT = 180^o - 150^o = 30^o$
$\angle PTQ = \angle RQT$
This implies,
$\angle RQT = \frac{30^o}{2} = 15^o$
$\angle TQR = 15^o$
Hence proved.
- Related Articles
- In the figure, $PQRS$ is a square and $SRT$ is an equilateral triangle. Prove that $PT = QT$."\n
- Prove that each angle of an equilateral triangle is $60^o$.
- In the figure, $\triangle ABC$ is an equilateral triangle. Find $m \angle BEC$."\n
- In the figure, if $ABC$ is an equilateral triangle. Find $\angle BDC$ and $\angle BEC$."\n
- In the figure, $O$ is the centre of the circle, prove that $\angle x = \angle y + \angle z$."\n
- In the figure, $p$ is a transversal to lines $m$ and $n, \angle 2 = 120^o$ and $\angle 5 = 60^o$. Prove that $m \parallel n$."\n
- In the figure, it is given that $RT = TS, \angle 1 = 2\angle 2$ and $\angle 4 = 2\angle 3$. Prove that: $\triangle RBT = \triangle SAT$."\n
- In the figure, $AB \parallel CD$ and $P$ is any point shown in the figure. Prove that:$\angle ABP + \angle BPD + \angle CDP = 360^o$"\n
- In a $\triangle ABC, \angle A = x^o, \angle B = 3x^o and \angle C = y^o$. If $3y - 5x = 30$, prove that the triangle is right angled.
- $O$ is the circumcentre of the triangle $ABC$ and $OD$ is perpendicular on $BC$. Prove that $\angle BOD = \angle A$.
- $PQRS$ is a rectangle. Find $\angle OPS$, if $\angle QOR$ is given. $( a).\ 42^o$ $( b).\ 56^o$."\n
- In the figure, $\triangle PQR$ is an isosceles triangle with $PQ = PR$ and $m \angle PQR = 35^o$. Find $m \angle QSR$ and $m \angle QTR$."\n
- In the figure, it is given that $O$ is the centre of the circle and $\angle AOC = 150^o$. Find $\angle ABC$."\n
- If the bisectors of the base angles of a triangle enclose an angle of $135^o$. Prove that the triangle is a right triangle.
- In the figure, $\angle 1 = 60^o$ and $\angle 2 = (\frac{2}{3})$rd a right angle. Prove that $l \parallel m$."\n
Kickstart Your Career
Get certified by completing the course
Get Started