Question

In the given figure (not to scale) O is the centre of the circle $\overline{AC}$ and $\overline{BD}$ intersect at P. PB = PC $\angle PBO={25}^{\circ }$ and $\angle BOC={130}^{\circ }$ then find $\angle ABP+\angle DCP$

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. None of these

Answer 1

Answer: (B)

${30}^{\circ }$

Solution:

Given that:

$\angle PBO={25}^{\circ }$ and $PB=PC$

To find:

$\angle ABP+\angle DCP=?$

Solution:

In $△OBC$

$OB=OC=r\therefore \angle OBC=\angle OCB$

where r is the radius of the circle.

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

or, $2\angle OBC={180}^{\circ }-{130}^{\circ }$

or, $\angle OBC={25}^{\circ }$

Also $PB=PC\therefore \angle PCO=\angle PBO={25}^{\circ }$

And

$2\angle BAC=\angle BOC$ (Angle made by any chord at the centre is twice of angle made at the circumference.)

or, $\angle BAC=\frac{{130}^{\circ }}{2}={65}^{\circ }$

Now,

In $△ABC$

$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$

or, $\angle ABP+\angle PBO+\angle OBC+\angle PCO+\angle OCB+\angle BAC={180}^{\circ }$

or, $\angle ABP+{25}^{\circ }+{25}^{\circ }+{25}^{\circ }+{25}^{\circ }+{65}^{\circ }={180}^{\circ }$

or, $\angle ABP={15}^{\circ }$

Now,

$\angle ABP=\angle DCP$ (Angles made by the same chord the circumference are equal)

or, $\angle ABP+\angle DCP={15}^{\circ }+{15}^{\circ }={30}^{\circ }$

Hence, B is the correct answer.