An exterior angle of a triangle is 105-0 and its two interior opposite angles are equal- Each of these equal angles is37dfrac-1-0-2-72dfrac-1-0-2-52dfrac-1-0-2-75-0

Question

Toppr Admin · Accepted Answer

Given- Exterior angle - 105-x2218-Let the two Interior angles be x-Sum of Interior opposite angles - Exterior anglex-x-1052x-105x-5212-x2218-

An exterior angle of a triangle is $105^{0}$ and its two interior opposite angles are equal. Each of these equal angles is
$37 \frac{1^{0}}{2}$
$72 \frac{1^{0}}{2}$
$52 \frac{1^{0}}{2}$
$75^{0}$

Given: Exterior angle = $105^{\circ}$
Let the two Interior angles be $x$ .
Sum of Interior opposite angles = Exterior angle
$x + x = 105$
$2 x = 105$
$x = 52 {\frac{1}{2}}^{\circ}$

An exterior angle of a triangle is 1050 and its two interior opposite angles are equal. Each of these equal angles is371027210252102750

Given: Exterior angle = 105∘Let the two Interior angles be x.Sum of Interior opposite angles = Exterior anglex+x=1052x=105x=5212∘

An exterior angle of a triangle is $105^{0}$ and its two interior opposite angles are equal. Each of these equal angles is
$37 \frac{1^{0}}{2}$
$72 \frac{1^{0}}{2}$
$52 \frac{1^{0}}{2}$
$75^{0}$

Given: Exterior angle = $105^{\circ}$
Let the two Interior angles be $x$ .
Sum of Interior opposite angles = Exterior angle
$x + x = 105$
$2 x = 105$
$x = 52 {\frac{1}{2}}^{\circ}$