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Circle — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsCircle3 Marks
Question
In the given figure, $\text{ABCD}$ is a cyclic quadrilateral whose diagonals intersect at $P$ such that $\angle\text{DBC}=60^\circ$ and $\angle\text{BAC}=40^\circ.$ Find
$i. \angle\text{BCD}$
$ii. \angle\text{CAD}$
✓

Answer

$i. \angle\text{BDC}=\angle\text{BAC}=40^\circ [$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$]$
$\Rightarrow\ \angle\text{BCD}+60^\circ+40^\circ=180^\circ$
$\Rightarrow\ \angle\text{BCD}=(180^\circ-100^\circ)=80^\circ$
$ii. \angle\text{CAD}=\angle\text{CBD} [$Angles in the same segment$]$
$=60^\circ$

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