Question

In the given circle with diameter AB, find the value of x.

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Answer 1

$\angle$ADB = ${90}^{\circ }$ [Angle in a semi-circle]

$\angle$ABD = $\angle$ACD = ${30}^{\circ }$ [$\angle$s on the same segment]

In $\Delta$ADB, we have

$\angle$BAD +$\angle$ADB +$\angle$ABD = ${180}^{\circ }$ (Sum of the $\angle$s of a $\Delta$ is ${180}^{\circ }$)

x + ${90}^{\circ }$ + ${30}^{\circ }$ = ${180}^{\circ }$

x = (${180}^{\circ }$ - ${120}^{\circ }$ ) = ${60}^{\circ }$

Hence, x = ${60}^{\circ }$.