Test Index

UPSC CDS 1 2023 Math Paper

Question Numbers: 99-100

Consider the following for the next two (02) items that follow

In the following figure, a triangle ABC is inscribed in a circle with centre at O. Let

\angle POA = x^{\circ}

and

\angle OQB = y^{\circ}

. Further,

OB = BQ

Question : 99 of 100

Marks: +1, -0

What is the relation between x and y ?

Solution:

Concept:

The exterior angle theorem and properties of an isosceles triangle are used.

Explanation:

Given

OB = BQ

, so

\triangle OBQ

is isosceles.

Thus,

\angle BOQ = \angle OQB = y^{\circ}

\angle OBA

is the exterior angle of

\triangle OBQ

So,

\angle OBA = \angle BOQ + \angle OQB = y^{\circ} + y^{\circ} = 2y^{\circ}

Since

OA = OB

(radii),

\triangle OAB

is isosceles.

Therefore,

\angle OAB = \angle OBA = 2y^{\circ}

\triangle AOB

\angle AOB = 180^{\circ} - 2y^{\circ} - 2y^{\circ} = 180^{\circ} - 4y^{\circ}

Also,

PQ

is a straight line, so

\angle POA + \angle AOB + \angle BOQ = 180^{\circ}

Hence,

x^{\circ} + \angle AOB + y^{\circ} = 180^{\circ}

⇒

\angle AOB = 180^{\circ} - x^{\circ} - y^{\circ}

Equating the two expressions for

\angle AOB

180^{\circ} - 4y^{\circ} = 180^{\circ} - x^{\circ} - y^{\circ}

Simplify to

x = 3y

Answer:

x = 3y

(Option C).

Go to Question:

Prev Question

Next Question