In the given figure, O is the centre of the circle. Where = x^ and = 3x^ are given. [figure in the original paper] (a) What is the sum of the opposite angles of a cyclic quadrilateral? Write it. [1]…

(a) Sum of the opposite angles of a cyclic quadrilateral

In a cyclic quadrilateral, the sum of the measures of a pair of opposite angles is 180°. This is known as the cyclic quadrilateral theorem.

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(b) Finding the value of $x$

From the figure, we observe the following:

• $\angle WXY=x^{\circ }$ is an inscribed angle subtending arc $WY$.
• $\angle WZY=3x^{\circ }$ is also an inscribed angle subtending the same arc $WY$.

Since both angles subtend the same arc, they must be equal: $$\angle WXY=\angle WZY$$ $$x=3x$$ This leads to a contradiction, so we must reconsider the configuration. Instead, note that $\angle WZY$ is an exterior angle for the cyclic quadrilateral $WXYZ$, and it is equal to the opposite interior angle $\angle WXY$: $$\angle WZY=\angle WXY+\angle WXY$$ But this is incorrect. Instead, we use the fact that the sum of opposite angles in a cyclic quadrilateral is $180^{\circ }$. Let’s denote the quadrilateral as $WXYZ$: $$\angle WXY+\angle WZY=180^{\circ }$$ $$x+3x=180^{\circ }$$ $$4x=180^{\circ }$$ $$x=\frac{180^{\circ }}{4}=45^{\circ }$$

Final value of $x$: 45°

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(c) Experimental verification of the circumference angle theorem

Objective: Verify that the circumference angle $\angle WXY$ is half of the central angle $\angle WOY$ subtending the same arc $WY$.

Procedure:

1. Draw two circles with radii of at least 3 cm each on a sheet of paper.
2. Mark the centers of the circles as $O_1$ and $O_2$.
3. On each circle, draw two points $W$ and $Y$ on the circumference such that the arc $WY$ is clearly visible.
4. Draw the chord $WY$ and measure the central angle $\angle W{O}_{1}Y$ using a protractor.
5. Draw a point $X$ on the circumference (not on the arc $WY$) and measure the inscribed angle $\angle WXY$.
6. Compare the two angles:
   • Central angle $\angle W{O}_{1}Y=2x$.
   • Circumference angle $\angle WXY=x$.

Observation: You will find that the inscribed angle $\angle WXY$ is always half of the central angle $\angle W{O}_{1}Y$ subtending the same arc $WY$. This confirms the theorem that the angle subtended by an arc at the center is twice the angle subtended at any point on the circumference.

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(d) Proof that $\angle OWZ+\angle OYZ=135^{\circ }$

From the figure, we know:

• $x=45^{\circ }$ (from part b).
• $\angle WXY=x=45^{\circ }$.
• $\angle WZY=3x=135^{\circ }$.

Since $O$ is the center, $\angle WOY$ is the central angle subtending arc $WY$. By the central angle theorem: $$\angle WOY=2\times \angle WXY=2\times 45^{\circ }=90^{\circ }$$

Now, consider triangles $OWZ$ and $OYZ$:

• In $△OWZ$, $\angle OWZ$ is half of $\angle WOY$ because $OZ$ is a radius and $WZ$ is a chord. However, we need to find $\angle OWZ$ directly.
• Similarly, $\angle OYZ$ is half of $\angle WOY$ if $Y$ lies on the same arc, but this is not the case here.

Instead, observe that:

• $\angle WZY=135^{\circ }$ is an exterior angle for $△OWZ$ at vertex $Z$.
• Therefore, $\angle WZY=\angle OWZ+\angle OYZ$.

But we need to find $\angle OWZ+\angle OYZ$. Since $OZ$ is a radius and $WZ$ and $YZ$ are chords, we use the fact that: $$\angle OWZ=\angle OYZ=\frac{1}{2}\angle WOY=\frac{1}{2}\times 90^{\circ }=45^{\circ }$$ However, this is incorrect because $\angle OWZ$ and $\angle OYZ$ are not necessarily equal.

Instead, let’s use the following approach:

• Since $O$ is the center, $OW=OZ=OY$ (all are radii).

• Thus, $△OWZ$ and $△OYZ$ are isosceles triangles.

• Let $\angle OWZ=\alpha$ and $\angle OYZ=\beta$.

• In $△OWZ$, the sum of angles is $180^{\circ }$: $$\angle WOZ+\angle OWZ+\angle OZW=180^{\circ }$$ Since $OW=OZ$, $\angle OWZ=\angle OZW=\alpha$. $$\angle WOZ+2\alpha =180^{\circ }$$ $$2\alpha =180^{\circ }-\angle WOZ$$

• Similarly, in $△OYZ$, since $OY=OZ$: $$\angle YOZ+\angle OYZ+\angle OZY=180^{\circ }$$ $$\angle YOZ+2\beta =180^{\circ }$$ $$2\beta =180^{\circ }-\angle YOZ$$

Now, note that $\angle WOZ+\angle YOZ=\angle WOY=90^{\circ }$. Therefore: $$2\alpha +2\beta =180^{\circ }-\angle WOZ+180^{\circ }-\angle YOZ$$ $$2(\alpha +\beta )=360^{\circ }-(\angle WOZ+\angle YOZ)$$ $$2(\alpha +\beta )=360^{\circ }-90^{\circ }=270^{\circ }$$ $$\alpha +\beta =135^{\circ }$$

Thus, we have proven that: $$\angle OWZ+\angle OYZ=135^{\circ }$$