Question

In the figure (ii) given below, I is the in centre of $\Delta ABC.$ $AI$ produced meets the circumcircle of $△ABC$ at $D.$ Given that $\angle ABC={55}^{\circ }$ and $\angle ACB={65}^{\circ }$

Calculate $\left(i\right)\angle BCD$ $\left(ii\right)\angle CBD$ $\left(iii\right)\angle DCI$ $\left(iv\right)\angle BIC$.

• A. ${50}^0,{20}^0,{32.5}^0,{240}^0$
• B. ${30}^0,{30}^0,{62.5}^0,{120}^0$
• C. ${35}^0,{35}^0,{32.5}^0,{100}^0$
• D. None of these

Answer 1

Answer: (B)

${30}^0,{30}^0,{62.5}^0,{120}^0$

Join $IA,IC$ and $CD$

(i) $IB$ is bisector of $\angle ABC$

$\angle =\frac{1}{2}\angle ABC=\frac{1}{2}(180-65-55)={30}^{\circ }$

Also $\angle DCA=\angle ABD={30}^{\circ }$

$\left(ii\right)\angle DAC=\angle CBD={30}^{\circ }$

$\left(ii\right)\angle ACI=\frac{\angle ACB}{2}=\frac{65}{2}={32.5}^{\circ }$

$\left(iv\right)\angle IAC=\frac{\angle BAC}{2}=\frac{55}{2}=27.5\circ$

$\angle AIC+\angle IAC+\angle ICA=180$

$\angle AIC=180-27.5-32.5={120}^{\circ }$