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 interior angles are $\angle ABC$,$\angle ACB$,and $\angle BAC$ (or $\angle A$).Since $\angle ABD$ is an exterior angle at vert

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) In $\Delta ABC$,the interior angles are $\angle ABC$,$\angle ACB$,and $\angle BAC$ (or $\angle A$).Since $\angle ABD$ is an exterior angle at vertex $B$,we have $\angle ABC + \angle ABD = 180^{\circ}$ (linear pair).$\angle ABC = 180^{\circ} - 110^{\circ} = 70^{\circ}$.Since $\angle ACE$ is an exterior angle at vertex $C$,we have $\angle ACB + \angle ACE = 180^{\circ}$ (linear pair).$\angle ACB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.$\angle A + 70^{\circ} + 30^{\circ} = 180^{\circ}$.$\angle A + 100^{\circ} = 180^{\circ}$.$\angle A = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD = 110^{\circ}$ and $\angle ACE = 150^{\circ}$,then $\angle A$ is: (in $^{\circ}$)

Similar Questions

$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle A = 50^{\circ}$ and $\angle B = 65^{\circ}$,then $\angle ACD = \dots$ (in $^{\circ}$)

In the figure,$PQ > PR$ and $QS$ and $RS$ are the bisectors of $\angle Q$ and $\angle R$,respectively. Show that $SQ > SR$.

In $\Delta ABC$,$AB = 6 \text{ cm}$ and $BC = 9 \text{ cm}$,then $AC < \dots \text{ cm}$.

Bisectors of the angles $B$ and $C$ of an isosceles triangle with $AB = AC$ intersect each other at $O$. $BO$ is produced to a point $M$. Prove that $\angle MOC = \angle ABC$.

In $\triangle ABC$,$AB = AC$ and $\angle B = 50^{\circ}$. Then $\angle C$ is equal to (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module