Compute the value of $x$ in each of the following figures:"

Given:

In $\triangle ABC$, sides $BC$ and $CA$ are produced to $D$ and $E$ respectively.

$\angle \mathrm{ACD}=112^{\circ}$ and $\angle \mathrm{BAE}=120^{\circ}$

To do:

We have to compute the value of $x$.

Solution:

$\angle \mathrm{ACB}+\angle \mathrm{ACD}=180^{\circ}$

$\angle \mathrm{ACB}+112^{\circ}=180^{\circ}$

$\angle \mathrm{ACB}=180^{\circ}-112^{\circ}=68^{\circ}$

Similarly,

$\angle \mathrm{BAE}+\angle \mathrm{BAC}=180^{\circ}$

$120^{\circ}+\angle \mathrm{BAC}=180^{\circ}$

$\angle \mathrm{BAC}=180^{\circ}-120^{\circ}=60^{\circ}$

$\angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{BCA}=180^{\circ}$

$60^{\circ}+x+68^{\circ}=180^{\circ}$

$128^{\circ}+x=180^{\circ}$

$x=180^{\circ}-128^{\circ}=52^{\circ}$

Hence, the value of $x$ is $52^{\circ}$.

Tutorialspoint

Updated on: 10-Oct-2022

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