Question

$ED$ is the tangent to the circle with centre $O.$ $\angle BCD={52}^{\circ }.$ Then, $\angle CAB$ equals

• A. ${38}^{\circ }$
• B. ${76}^{\circ }$
• C. ${52}^{\circ }$
• D. ${46}^{\circ }$

Answer 1

Answer: (C)

${52}^{\circ }$

Given, $\angle BCD={52}^{\circ }$
Now, $\angle OCD={90}^{\circ }$ ....(Right angle is formed between theradius an the tangent a the point on circle)
$\Rightarrow \angle BCD+\angle OCB=90$
$\Rightarrow \angle OCB=90-52={38}^{\circ }$
Now, in $△OCB$, we have
$OB=OC$
$\Rightarrow \angle OCB=\angle OBC={38}^{\circ }$ ....(Isosceles triangle property)
Sum of angles of triangle $={180}^o$
$\Rightarrow \angle OCB+\angle OBC+\angle COB=180$
$\Rightarrow 38+38+\angle COB=180$
$\Rightarrow \angle COB=180-76$
$\Rightarrow \angle COB={104}^{\circ }$
Now, $\angle CAB=\frac{1}{2}\angle COB=\frac{1}{2}\left(104\right)={52}^{\circ }$
Angle formed at the circle is half the angle subtended at the center by the segment.