angle ABD and angle ACE are exterior angles o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In $\Delta ABC$,the exterior angle at vertex $B$ is $\angle ABD = 140^{\circ}$. Since the sum of an interior angle and its adjacent exterior angle

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) In $\Delta ABC$,the exterior angle at vertex $B$ is $\angle ABD = 140^{\circ}$. Since the sum of an interior angle and its adjacent exterior angle is $180^{\circ}$,we have $\angle ABC = 180^{\circ} - 140^{\circ} = 40^{\circ}$.Similarly,the exterior angle at vertex $C$ is $\angle ACE = 80^{\circ}$. Thus,the interior angle $\angle ACB = 180^{\circ} - 80^{\circ} = 100^{\circ}$.In $\Delta ABC$,the sum of all interior angles is $180^{\circ}$. Therefore,$\angle A + \angle ABC + \angle ACB = 180^{\circ}$.Substituting the values,$\angle A + 40^{\circ} + 100^{\circ} = 180^{\circ}$.$\angle A + 140^{\circ} = 180^{\circ}$.$\angle A = 180^{\circ} - 140^{\circ} = 40^{\circ}$.

$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD = 140^{\circ}$ and $\angle ACE = 80^{\circ}$,then find $\angle A$. (in $^{\circ}$)

Similar Questions

If in $\Delta XYZ$,$\angle Y > \angle X > \angle Z$,then find the smallest side of the triangle.

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.

In $\Delta ABC$,$AB = 8 \, cm$ and $BC = 5 \, cm$,then $AC > \ldots \ldots \ldots cm$.

In a triangle $ABC$,$D$ is the mid-point of side $AC$ such that $BD = \frac{1}{2} AC$. Show that $\angle ABC$ is a right angle.

In $\Delta PQR$,$S$ is any point on the side $QR$. Prove that $PQ + QR + RP > 2PS$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module