Question

In the above figure O is the center of the circle and $\angle$AOB = ${80}^{\circ }$, then complete the matching.

$\begin{array}{cc}\text{Angle}& \text{Value}\\ \text{1.ACB}& \text{a.}{90}^0\\ \text{2.ADB}& \text{b.}{40}^0\\ \text{3.ABC}& \text{c.}{80}^0\\ \text{4.AOC}& \text{d.}{180}^0\end{array}$

• A. 1- a, 2- d, 3-c, 4 -b
• B. 1- a, 2- c, 3-d, 4 -b
• C. 1- b, 2- b, 3-a, 4 -d
• D. 1- b, 2- a, 3-c, 4 -d

Answer 1

The correct option is C

1- b, 2- b, 3-a, 4 -d

AOC is a straight line, so $\angle$AOC should be a straight angle. Therefore $\angle$AOC= ${180}^{\circ }$.

Diameter is also a chord of the circle. The angle subtended by AC at the center is $\angle$AOC = ${180}^{\circ }$

Since angle subtended by a chord at the center is equal to twice the angle subtended by chord at any point on the circle,

We get $\angle$AOC = 2 $\angle$ABC

$\angle$ABC = ${90}^{\circ }$

Similarly, $\angle$ADB = $\frac{1}{2}$ $\angle$AOB

$\angle$ACB= $\frac{1}{2}$ $\angle$AOB

From the above equations we can see that $\angle$ADB = $\angle$ACB = $\frac{1}{2}$ $\angle$AOB

$\angle$ADB = $\angle$ACB = ${40}^{\circ }$