Question

If O is the centre of the circle. Find the value of $x{}^\circ $ in the following figure:

Answer 1

Hint: We can solve this problem by using two concepts which are,” Angles inscribed by the same arc on the circumference of a circle are always EQUAL.” And “Semicircular Angles are always Right angles.”

Complete step-by-step answer:
Firstly, we will write the given values from figure,
$\angle ABC=40{}^\circ ,\angle BDC=x{}^\circ $………………………………….. (1)
To find the $\angle BDC$ we should know the key concept given below,
Concept: Angles inscribed by the same arc on the circumference of a circle are always EQUAL.
i.e. $\angle BAC=\angle BDC$…………………………………. (2)
As O is the centre of the circle, AB is the diameter, which can be easily seen from the figure and therefore $\angle ACB$ is a semicircular angle.
As we know Semicircular Angles are always Right angles.
Therefore, $\angle ACB=90{}^\circ $………………………………… (3)
Consider $\Delta ABC$,

$\angle ACB+\angle ABC+\angle BAC=180{}^\circ $ (Angles of a Triangle)
$\therefore 90{}^\circ +40{}^\circ +\angle BAC=180{}^\circ $ [From (1) and (3)]
$\therefore \angle BAC=180{}^\circ -130{}^\circ $
$\therefore \angle BAC=50{}^\circ $
Now, To find the value of $\angle BDC$ rewrite the equation (2) which we have evaluated earlier,
$\angle BAC=\angle BDC$
Put, $\angle BAC=50{}^\circ $
$\therefore \angle BDC=50{}^\circ $
If we refer Equation (1) then we can write,
$x{}^\circ =\angle BDC$
$\therefore x{}^\circ =\angle BDC=50{}^\circ $
$\therefore $ The value of $x{}^\circ $ in the following figure is $50{}^\circ $.

Note: If we see the geometry of figure, we can easily calculate value of $x{}^\circ $ by considering the diameters perpendicular to each other and therefore considering the angle B to be $90{}^\circ $ and proceeding further for solution, but don’t proceed like this as condition is not mentioned in the problem and there is no sign of showing perpendicular in the figure.