Question

Using Central angle theorem 10 th standard​

Answer 1

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Answer 2

(a) Steps of construction:

1. Draw a circle with centre O and having radius 3 cm.

2. To make the three points A, B, and C in the circle, join A to the centre of the circle O.

3. If m∠BAC is to be 50°, m∠BOC should be 100°

4. If m∠ABC is to be 60°, m∠AOC should be 120°

5. If m∠ACB is to be 70°, m∠AOB should be 140°

6. Join the points B and C such that, m∠AOC=100° and m∠BOC=120°

Thus, ΔABC is the required triangle

(b) Steps of construction:

1. Extend OA and draw perpendicular to it through A.

2. Extend OB and draw perpendicular to it through B.

3. In the same way draw perpendicular to OC through C.

Let the points of intersection of these perpendicular be P, Q and R, so we get the required ΔPQR.

(c) In the quadrilateral PAOC,

m∠AOC=120°

⇒m∠P=180°−120°=60°

(opposite angles of a quadrilateral are suplementary)

In the same way, m∠Q=180°−140°=40°

And, m∠R=180°−100° =80°