Question

In the given figure, AB is a diameter of the circle and points C and D are on the circumference such that $\angle CAD={30}^o$ and $\angle CBA={70}^o$ what is the measure of $\angle ACD$?

• A. ${40}^o$
• B. ${50}^o$
• C. ${30}^o$
• D. ${90}^o$

Answer 1

The correct option is A ${40}^o$

$\angle$ACB=90$°$ (Angle subtended on semi-circle)

In, $△$ACB,

$\angle$CAB=180$°$-($\angle$ACB+$\angle$CBA)

=180$°$-(90$°$+70$°$)

=20$°$

Now,

$\angle$DCA+$\angle$ACB+$\angle$BAC+$\angle$DAC=180$°$ [Sum of opp. anglesof cyclic quadrilateralis supplementary.]

$\angle$DCA+90$°$+20$°$+30$°$=180$°$

$\angle$ACD=180$°$-40$°$

$\angle$ACD=40$°$

Hence, 40$°$ is the correct answer.