Question
Find $x, y, z$ (whichever is required) from the figure given below:
✓
Answer
We can see that in $\triangle \text{ADC}, \angle \text{ADC}$ is equal to $90^\circ $
$(\triangle \text{ADC}$ is a right triangle$)$
We also know that the sum of all the angles of a triangle is equal to $180^\circ $
Which means: $45^\circ + 90^\circ + y = 180^\circ $
$($Sum of the angles of $\triangle \text{ADC})$
$135^\circ + y = 180^\circ $
$y = 180^\circ - 135^\circ $
$y = 45^\circ $
We can also say that in $\triangle \text{ABC},\angle \text{ABC}+\angle \text{ACB}+\angle \text{BAC}$ is equal to $180^\circ $
$($Sum of the angles of $\triangle \text{ABC})$
$40^\circ + y + (x + 45^\circ ) = 180^\circ $
$40^\circ + 45^\circ + x + 45^\circ = 180^\circ (y = 45^\circ )$
$x = 180^\circ - 130^\circ $
$x = 50^\circ $
Therefore, we can say that the required angles are $45^\circ $ and $50^\circ $
$(\triangle \text{ADC}$ is a right triangle$)$
We also know that the sum of all the angles of a triangle is equal to $180^\circ $
Which means: $45^\circ + 90^\circ + y = 180^\circ $
$($Sum of the angles of $\triangle \text{ADC})$
$135^\circ + y = 180^\circ $
$y = 180^\circ - 135^\circ $
$y = 45^\circ $
We can also say that in $\triangle \text{ABC},\angle \text{ABC}+\angle \text{ACB}+\angle \text{BAC}$ is equal to $180^\circ $
$($Sum of the angles of $\triangle \text{ABC})$
$40^\circ + y + (x + 45^\circ ) = 180^\circ $
$40^\circ + 45^\circ + x + 45^\circ = 180^\circ (y = 45^\circ )$
$x = 180^\circ - 130^\circ $
$x = 50^\circ $
Therefore, we can say that the required angles are $45^\circ $ and $50^\circ $
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