Question

In the given figure, FDE is the tangent to the circle at point D. If $\angle DAB={56}^{\circ }and\angle DBC={30}^{\circ }$, then the measure of $\angle CDF+\angle BDC$ is equal to

• A.
  ${124}^{\circ }$
• B.
  ${134}^{\circ }$
• C.
  
  ${176}^{\circ }$
• D.
  ${156}^{\circ }$

Answer 1

The correct option is C


${176}^{\circ }$
Given: $\angle DAB={56}^{\circ }and\angle DBC={30}^{\circ }$



Since, FDE is tangent to circle and ABCD is a cyclic quadrilateral.

Now, $\angle BAD+\angle BCD={180}^{\circ }$
(Sum of opposite angles in a cyclic quadrilateral)
$\Rightarrow \angle BCD={180}^{\circ }-{56}^{\circ }\left(Given\right)\Rightarrow \angle BCD={124}^{\circ }....\left(i\right)$

In $\Delta BCD,\angle CBD+\angle BCD+\angle BDC={180}^{\circ }$

(By Angle sum property)

$\Rightarrow {124}^{\circ }+{30}^{\circ }+\angle BDC={180}^{\circ }\Rightarrow \angle BDC={180}^{\circ }-{154}^{\circ }\Rightarrow \angle BDC={26}^{\circ }$

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

$\therefore \angle BOD=2\angle BAD={112}^{\circ }\dots ..\left(iii\right)Since,OB=OD.\left(Radii\right)\Rightarrow \angle OBD=\angle ODB.....\left(iv\right)\text{(Angle opp. to equal sides are}\text{always equal)}In\Delta BOD,ByAnglesumproperly,\angle BOD+\angle ODB+\angle OBD={180}^{\circ }\Rightarrow {112}^{\circ }+\angle ODB+\angle ODB={180}^{\circ }\left[From\right(iii\left)and\right(iv\left)\right]\Rightarrow 2\angle ODB=180º–{112}^{\circ }\Rightarrow \angle ODB=\frac{{180}^{\circ }-{112}^{\circ }}{2}={34}^{\circ }....\left(v\right)Since,OD\perp FE,\angle ODF={90}^{\circ }.$

(Angle between a tangent and the radius of a circle is ${90}^{\circ }$)
$\therefore \angle CDF+BDC=\angle ODF+\angle ODB+2\angle BDC={90}^{\circ }+{34}^{\circ }+{52}^{\circ }\left[from\right(ii\left)and\right(v\left)\right]={176}^{\circ }$

Hence, the correct answer is option c .