Advertisements
Advertisements
Question
In the following figure, if ∠OAB = 40º, then ∠ACB is equal to ______.
Options
50º
40º
60º
70°
Advertisements
Solution
In the following figure, if ∠OAB = 40º, then ∠ACB is equal to 50º.
Explanation:
Given: ∠OAB = 40º
Now, in triangle OAB,
OA = OB ...[Radii of circle]
So, ∠OAB = ∠OBA = 40º ...[Angle opposite to equal sides are equal]
So, ∠AOB = 180º – (40º + 40º) = 100º
As we know that angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
So, ∠ACB = `1/2` ∠AOB = `1/2 xx 100^circ` = 50º
APPEARS IN
RELATED QUESTIONS
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.
In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F Respectively. If AB= 12cm, BC=8cm and AC = 10cm, find the length of AD, BE and CF.
In a cyclic quadrilateral ABCD if AB || CD and ∠B = 70°, find the remaining angles.
A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm2, find the sides PQ and PR.
In the given figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ΔAEO ∼ Δ ABC.
In Fig. 8.79, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2.
Find the area of a circle of radius 7 cm.
Mark two points A and B ,4cm a part, Draw a circle passing through B and with A as a center
The diameter of a circle is 12.6 cm. State, the length of its radius.
Three circles touch each other externally. The distance between their centres is 5 cm, 6 cm, and 7 cm. Find the radii of the circles.
