In the following figure,in Delta ABC,angle AB | vedclass

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

Question

(N/A) Given that $\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$.Since $DBC$ is a straight line,$\angle ABC$ and $\angle ABD$ form a linear pa

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Given that $\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$.
Since $DBC$ is a straight line,$\angle ABC$ and $\angle ABD$ form a linear pair.
Therefore,$\angle ABD + \angle ABC = 180^{\circ}$.
$\angle ABD + 70^{\circ} = 180^{\circ} \implies \angle ABD = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
Similarly,$\angle ACB$ and $\angle ACE$ form a linear pair on the straight line $BCE$.
Therefore,$\angle ACE + \angle ACB = 180^{\circ}$.
$\angle ACE + 60^{\circ} = 180^{\circ} \implies \angle ACE = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,$\angle ABD = 110^{\circ}$ and $\angle ACE = 120^{\circ}$.

In the following figure,in $\Delta ABC$,$\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$. Find $\angle ABD$ and $\angle ACE$.

Similar Questions

In $\Delta ABC$,$\angle A = 2x - 10^{\circ}$,$\angle B = x + 10^{\circ}$,and $\angle C = 2x - 20^{\circ}$. Find the measure of each angle of $\Delta ABC$.

$\angle ACD$ is an exterior angle of $\Delta ABC$. The bisector of $\angle ABC$ and exterior $\angle ACD$ intersect each other at point $E$. Prove that,$\angle BEC = \frac{1}{2} \angle BAC$.

Prove that through a given point,we can draw only one perpendicular to a given line.

In the figure,$m$ and $n$ are two plane mirrors perpendicular to each other. Show that the incident ray $CA$ is parallel to the reflected ray $BD$.

Ray $YM$ is the bisector of $\angle XYZ$ and ray $YN$ is the bisector of $\angle MYZ$. If $\angle XYN = 45^{\circ}$,then find $\angle XYZ$. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module