Question

In the adjoining figure A, D, B, C are four points on the circumference of a circle with centre O. Arc AB = 2 Arc BC and $\angle$AOB = ${108}^0$. FInd:

i. $\angle$ACB

ii. $\angle$CAB

III. $\angle$ADB

Answer 1

Let $\angle$BOC = $x^{\circ }$. Then $\angle$AOB = $2x^{\circ }$

Now, $2x^{\circ }$ = ${108}^{\circ }$

$\Rightarrow$ x = ${54}^{\circ }$

$\therefore$ $\angle$BOC = ${54}^{\circ }$ and $\angle$AOB = ${108}^{\circ }$

i. $\angle$ACB = $\frac{1}{2}$$\angle$AOB = $\frac{1}{2}$ $\times$ ${108}^{\circ }$ = ${54}^{\circ }$ (since angle at centre is double the angle at a point on the circumference)

ii. $\angle$CAB = $\frac{1}{2}$$\angle$COB = $\frac{1}{2}$$\angle$BOC = $\frac{1}{2}$ $\times$ ${54}^{\circ }$ = ${27}^{\circ }$

iii. $\angle$ADB = $\frac{1}{2}$ $\angle$AOB = $\frac{1}{2}$ $\times$ [${360}^{\circ }$ - ${108}^{\circ }$] = ${126}^{\circ }$