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CAT 2017 Slot 2 — QA Question 18

Basics of QuadrilateralsEasy

ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is

Answer & solution

Correct answer: 90

Answer: 90

Solution

Easy

The central angle ∠COD\angle COD over arc CDCD is twice the inscribed angle ∠CAD\angle CAD over the same arc. Add ∠BAC\angle BAC to get ∠BAD\angle BAD, then use the cyclic-quadrilateral property that opposite angles sum to 180∘180^\circ.

O A C D B 120°
1

Inscribed angle on arc CD. The angle at the centre is twice the inscribed angle subtending the same arc, so ∠CAD\angle CAD is half of ∠COD\angle COD.

∠CAD=12 ∠COD=12(120∘)=60∘\begin{aligned} &\angle CAD = \tfrac12\,\angle COD = \tfrac12(120^\circ) = 60^\circ \end{aligned}
2

Angle BAD. Vertex AA sees B,C,DB,C,D in order, so ∠BAD=∠BAC+∠CAD\angle BAD = \angle BAC + \angle CAD.

∠BAD=30∘+60∘=90∘(given ∠BAC=30∘, step 1)\begin{aligned} &\angle BAD = 30^\circ + 60^\circ = 90^\circ \quad\text{(given } \angle BAC=30^\circ\text{, step 1)} \end{aligned}
3

Opposite angles of a cyclic quadrilateral. In ABCDABCD the angles ∠BAD\angle BAD and ∠BCD\angle BCD are opposite, so they sum to 180∘180^\circ.

∠BAD+∠BCD=180∘⇒ ∠BCD=180∘−90∘=90∘(from step 2)\begin{aligned} &\angle BAD + \angle BCD = 180^\circ\\ &\Rightarrow\ \angle BCD = 180^\circ - 90^\circ = 90^\circ \quad\text{(from step 2)} \end{aligned}
∠BCD=90∘\angle BCD = 90^\circ

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CAT 2017 Slot 2 QA Q18 — Basics of Quadrilaterals solution | TheCATExam