Question

In the figure, given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given $\angle XTY={80}^{\circ }$ and $\angle XOZ={140}^{\circ }$, calculate the value of $\angle ZXY$.

Answer 1

In the figure, a circle with centre O, is the circumcircle of triangle XYZ.

$\angle XOZ={140}^{\circ }$ (given)

Tangents from X and Z to the circle meet at T such that $\angle XTY={80}^{\circ }$

Note that since a tangent at any point of a circle is perpendicular to the radius at the point of contact, we have $\angle OXT=\angle OYT={90}^{\circ }$ (1 mark)

Since sum of angles of a quadrilateral is 360 degrees,$\angle XOY={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{80}^{\circ })$
$={360}^{\circ }-{260}^{\circ }={100}^{\circ }$

But $\angle XOY+\angle YOZ+\angle XOZ={360}^{\circ }$

(Angles at a point)

$\Rightarrow {100}^{\circ }+\angle YOZ+{140}^{\circ }={360}^{\circ }$

$\Rightarrow \angle YOZ={360}^{\circ }-{240}^{\circ }={120}^{\circ }$ ( 1 mark)

Now arc YZ subtends $\angle YOZ$ at the centre are $\angle YXZ$ at the remaining part of the circle.

$\therefore \angle YOZ=2\angle YXZ$

$\Rightarrow \angle YXZ=\frac{1}{2}\angle YOZ=\frac{1}{2}\times {120}^{\circ }$
Thus, $\angle YXZ={60}^{\circ }$ ( 1 marks)