Question
Find $x, y, z$ (whichever is required) from the figure given below:
✓
Answer
We know that the sum of all the angles of a triangle is equal to $180^\circ$
Therefore, for $\triangle \text{ABD}:$
$\angle \text{ABD}+\angle \text{ADB}+\angle \text{BAD}=180^\circ$
$($Sum of the angles of $\triangle \text{ABD})$
$50^\circ + x + 50^\circ = 180^\circ 100^\circ + x = 180^\circ x = 180^\circ - 100^\circ x = 80^\circ $
For $\triangle \text{ABC}:$
$\angle \text{ABC}+\angle \text{ACB}+\angle \text{BAC}=180^\circ$
$($Sum of the angles of $\triangle \text{ABC})$
$50^\circ + z + (50^\circ + 30^\circ ) = 180^\circ 50^\circ + z + 50^\circ + 30^\circ = 180^\circ z = 180^\circ - 130^\circ z = 50^\circ $
Using the same argument for $\triangle \text{ADC}:$
$\angle \text{ADC}+\angle \text{ACD}+\angle \text{DAC}=180^\circ$
$($Sum of the angles of $\triangle \text{ADC})$
$y + z + 30^\circ = 180^\circ y + 50^\circ + 30^\circ = 180^\circ (z = 50^\circ ) y = 180^\circ - 80^\circ y = 100^\circ $
Therefore, we can conclude that the required angles are $80^\circ , 50^\circ $ and $100^\circ $
Therefore, for $\triangle \text{ABD}:$
$\angle \text{ABD}+\angle \text{ADB}+\angle \text{BAD}=180^\circ$
$($Sum of the angles of $\triangle \text{ABD})$
$50^\circ + x + 50^\circ = 180^\circ 100^\circ + x = 180^\circ x = 180^\circ - 100^\circ x = 80^\circ $
For $\triangle \text{ABC}:$
$\angle \text{ABC}+\angle \text{ACB}+\angle \text{BAC}=180^\circ$
$($Sum of the angles of $\triangle \text{ABC})$
$50^\circ + z + (50^\circ + 30^\circ ) = 180^\circ 50^\circ + z + 50^\circ + 30^\circ = 180^\circ z = 180^\circ - 130^\circ z = 50^\circ $
Using the same argument for $\triangle \text{ADC}:$
$\angle \text{ADC}+\angle \text{ACD}+\angle \text{DAC}=180^\circ$
$($Sum of the angles of $\triangle \text{ADC})$
$y + z + 30^\circ = 180^\circ y + 50^\circ + 30^\circ = 180^\circ (z = 50^\circ ) y = 180^\circ - 80^\circ y = 100^\circ $
Therefore, we can conclude that the required angles are $80^\circ , 50^\circ $ and $100^\circ $
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