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°
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° + x + 50° = 180°
100° + x = 180°
x = 180° - 100°
x = 80°
For $\triangle \text{ABC}:$
$\angle \text{ABC}+\angle \text{ACB}+\angle \text{BAC}=180^\circ$ $($Sum of the angles of $\triangle \text{ABC})$
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° - 130°
z = 50°
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° = 180°
y + 50° + 30° = 180° (z = 50°)
y = 180° - 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°
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° + x + 50° = 180°
100° + x = 180°
x = 180° - 100°
x = 80°
For $\triangle \text{ABC}:$
$\angle \text{ABC}+\angle \text{ACB}+\angle \text{BAC}=180^\circ$ $($Sum of the angles of $\triangle \text{ABC})$
50° + z + (50° + 30°) = 180°
50° + z + 50° + 30° = 180°
z = 180° - 130°
z = 50°
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° = 180°
y + 50° + 30° = 180° (z = 50°)
y = 180° - 80°
y = 100°
Therefore, we can conclude that the required angles are 80°, 50° and 100°
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