Exterior Angle of a Triangle and Its Property

In the given figure, find $\angle 1$.

Find the value of $x$ in given figure.

In the following figure, find $x$. (in degree)

The exterior angles of a triangle is $125°$, if one of its interior opposite angle is $60°$, find the remaining angles of the triangle.

In the figure, find the measure of $x$:

In a triangle $PQR$ , the sides $PQ$ and $PR$ are produced to points $E$ and $D$ respectively. If bisectors $QO$ and $RO$ of angle $RQE$ and angle $QRD$ respectively meet at point O, then prove that: $\angle QOR=90°-\frac{1}{2}\angle QPR$.

An exterior angle of a triangle is of measure $80°$ and one of its interior opposite angles is of measure $30°$. Find the measure of the other interior opposite angle. 

An exterior angle of a triangle is of measure $70°$ and one of its interior opposite angles is of measure $25°$. Find the measure of the other interior opposite angle. 

Find the value of $x°, y°$ and $z°$

What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle?

In the given figure, $\mathrm{QAP}$ is the tangent at point $\mathrm{A}$ and $\mathrm{PBD}$ is a straight line. If $\angle \mathrm{ACB}=36º$ and $\angle \mathrm{APB}=42º,$ find $\angle \mathrm{QAD}$.

In the given figure, $\mathrm{QAP}$ is the tangent at point $\mathrm{A}$ and $\mathrm{PBD}$ is a straight line. If $\angle \mathrm{ACB}=36º$ and $\angle \mathrm{APB}=42º,$ find $\angle \mathrm{ABD}$.

In the given figure, if $x=k°$ then, find the value of $k$.

Find the perimeter of the given figures, if the sides of the triangle are given in the lateral figure.

Where $a=3 \mathrm{cm}, b=2 \mathrm{cm}$

By how much is $\angle \mathrm{a}$ bigger than $\angle \mathrm{b}$ in the given figure.

The sides of a triangle are produced in order. The sum of the three exterior angles so formed is :-

Find $\mathrm{x}$ in the figure shown along side.

In figure,$AB=AC,\angle ACM$ $=125°$ and $\angle PAB=x.$ Find the value of $x$.

In quadrilateral $ABCD$, $AB=AD$ and $BC=CD$. Show that $\angle ABC=\angle ADC$.

$D$ is a point on side $BC$ of $\Delta ABC$ such that $AD=AC$ (see figure). Show that $AB>AD$.