Question

Question 129

In the given figure, find the values of x, y and z.

Answer 1

In the given figure, $\angle BAD={60}^{\circ },\angle ABD={60}^{\circ },\angle ADB=x,\angle DAC={30}^{\circ },\angle ADC=y$ and $\angle ACD=z$
We know that, the sum of all angles in a triangle is equal to ${180}^{\circ }$
In $\Delta ABD,\angle BAD+\angle ABD+\angle ADB={180}^{\circ }$
$\Rightarrow {60}^{\circ }+{60}^{\circ }+x={180}^{\circ }\Rightarrow {120}^{\circ }+x={180}^{\circ }\Rightarrow x={180}^{\circ }-{120}^{\circ }\Rightarrow x={60}^{\circ }$
Now, $y=\angle BAD+\angle ABD$
[$\because$ exterior angle is equal to the sum of interior opposite angles]
$\Rightarrow y={60}^{\circ }+{60}^{\circ }\therefore y={120}^{\circ }$
In $\Delta ADC,\angle DAC+\angle ADC+\angle ACD={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow {30}^{\circ }+{120}^{\circ }+z={180}^{\circ }\Rightarrow {150}^{\circ }+z={180}^{\circ }\Rightarrow z={180}^{\circ }-{150}^{\circ }\Rightarrow z={30}^{\circ }$
Hence, $x={60}^{\circ },y={120}^{\circ }$ and $z={30}^{\circ }$