Question
Find $x, y, z$ (whichever is required) from the figure given below:
✓
Answer
In $\triangle \text{ABC}$ and $\triangle \text{ADE}$
we have: $\angle \text{ADE}=\angle \text{ABC}$ (Corresponding angles) $\text{x}=40^\circ$
$\angle \text{AED}=\angle \text{ACB}$ (Corresponding angles) $\text{y}=30^\circ$
We know that the sum of all the three angles of a triangle is equal to $180^\circ x + y + z = 180^\circ $
$($Angles of $\triangle \text{ADE})$
Which means: $40^\circ + 30^\circ + z = 180^\circ z = 180^\circ - 70^\circ z = 110^\circ $
Therefore, we can conclude that the three angles of the given triangle are $40^\circ , 30^\circ $ and $110^\circ $
we have: $\angle \text{ADE}=\angle \text{ABC}$ (Corresponding angles) $\text{x}=40^\circ$
$\angle \text{AED}=\angle \text{ACB}$ (Corresponding angles) $\text{y}=30^\circ$
We know that the sum of all the three angles of a triangle is equal to $180^\circ x + y + z = 180^\circ $
$($Angles of $\triangle \text{ADE})$
Which means: $40^\circ + 30^\circ + z = 180^\circ z = 180^\circ - 70^\circ z = 110^\circ $
Therefore, we can conclude that the three angles of the given triangle are $40^\circ , 30^\circ $ and $110^\circ $
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