Radius=3cm and Angles=40 degree and 62½ degree
Answers
Answer:
(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 50o,m∠BOC should be 100o.
4. If m∠ABC is to be 60o,m∠AOC should be 120o.
5. If m∠ACB is to be 70o,m∠AOB should be 140o.
6. Join the points B and C such that m∠AOC=100o and m∠BOC=120o>
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=120o
⇒m∠P=180o−120o=60o .... (opposite angles of a quadrilateral are suplementary)
In the same way, m∠Q=180o−140o=40o
And, m∠R=180o−100o=80o