In ABC, D is a point in AB such that AC = CD = DB. If B = 28^ , find the ACD.

DBC is an isosceles triangle. A s, side C D= Side DB DBC= DCB [lf two sides of a triangle are equal, then angles opposite to them are equal] And B= DBC= DCB=28^ As the sum of all the angles of the triangle is 180^ DCB+ DBC+ BCD=180^ 28^ +28^ + BCD=180^ BCD=180^ -56^ BCD=124^ Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle. DBC+ DCB= ADC 28^ +28^ = ADC ADC=56^ Now ACD is an isosceles triangle with AC=DC ADC= DAC=56^ Sum of all the angles of a triangle is 180^ ACD+ ADC+ DAC=180^ ACD+56^ +56^ =180^ ACD=180^ -112^ ACD=68^

In ABC, D is a point in AB such that AC = CD = DB. If B = 28^ , f…

Download App

Question

In $\triangle ABC, D$ is a point in $AB$ such that $AC = CD = DB$. If $\angle B = 28^\circ $, find the $\angle ACD.$

✓

Answer

$\triangle \mathrm{DBC}$ is an isosceles triangle.
A s, side $C D=$ Side $DB$
$\Rightarrow \angle \mathrm{DBC}=\angle \mathrm{DCB} \ldots[$lf two sides of a triangle are equal, then angles opposite to them are equal$]$
And $\angle \mathrm{B}=\angle \mathrm{DBC}=\angle \mathrm{DCB}=28^{\circ}$
As the sum of all the angles of the triangle is $180^{\circ}$
$\angle \mathrm{DCB}+\angle \mathrm{DBC}+\angle \mathrm{BCD}=180^{\circ}$
$ \Rightarrow 28^{\circ}+28^{\circ}+\angle \mathrm{BCD}=180^{\circ}$
$ \Rightarrow \angle \mathrm{BCD}=180^{\circ}-56^{\circ}$
$ \Rightarrow \angle \mathrm{BCD}=124^{\circ}$
Sum of two non$-$adjacent interior angles of a triangle is equal to the exterior angle.
$\Rightarrow \angle \mathrm{DBC}+\angle \mathrm{DCB}=\angle \mathrm{ADC}$
$ \Rightarrow 28^{\circ}+28^{\circ}=\angle \mathrm{ADC}$
$ \Rightarrow \angle \mathrm{ADC}=56^{\circ}$
Now $\triangle \mathrm{ACD}$ is an isosceles triangle with $\mathrm{AC}=\mathrm{DC}$
$\Rightarrow \angle \mathrm{ADC}=\angle \mathrm{DAC}=56^{\circ}$
Sum of all the angles of a triangle is $180^{\circ}$
$\Rightarrow \angle \mathrm{ACD}+\angle \mathrm{ADC}+\angle \mathrm{DAC}=180^{\circ}$
$ \Rightarrow \angle \mathrm{ACD}+56^{\circ}+56^{\circ}=180^{\circ}$
$ \Rightarrow \angle \mathrm{ACD}=180^{\circ}-112^{\circ}$
$ \Rightarrow \angle \mathrm{ACD}=68^{\circ}$