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Circles — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsCircles3 Marks
Question
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that $\angle\text{DBC}=60^\circ$and $\angle\text{BAC}=40^\circ.$ Find
  1. $\angle\text{BCD}$
  2. $\angle\text{CAD}$
✓

Answer

  1. $\angle\text{BDC}=\angle\text{BAC}=40^\circ$ [Angles in the same segment]
In $\triangle\text{BCD},$ we have:

$\angle\text{BCD}+\angle\text{DBC}+\angle\text{BDC}=180^\circ$ [Angle sum property of a triangle]

$\Rightarrow\ \angle\text{BCD}+60^\circ+40^\circ=180^\circ$

$\Rightarrow\ \angle\text{BCD}=(180^\circ-100^\circ)=80^\circ$
  1. $\angle\text{CAD}=\angle\text{CBD}$ [Angles in the same segment]
$=60^\circ$

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