Prove That : No tangent can be drawn from the point `(5/2,1)` to the circumcircle of the triangle with vertices `(1,sqrt(3)),(1,-sqrt(3)),(3,-sqrt(3))` .

Show that the number of tangents that can be drawn from the point (5/2,1) to the circle passing through the points (1,sqrt(3)),(1,-sqrt(3)) and (3,-sqrt(3)) is zero.

The incentre of a triangle with vertices (7, 1),(-1, 5) and (3+2sqrt(3),3+4sqrt(3)) is

The centroid of the triangle with vertices (1, sqrt(3)), (0, 0) and (2, 0) is

(sqrt(3)+1+sqrt(3)+1)/(2sqrt(2))

The orthocentre of triangle with vertices (2,(sqrt3-1)/2) , (1/2,-1/2) , (2,,-1/2)

The orthocentre of triangle with vertices (2,(sqrt3-1)/2) , (1/2,-1/2) , (2,,-1/2)

The distance between the orthocentre and circumcentre of the triangle with vertices (1,2),\ (2,1)\ a n d\ ((3+sqrt(3))/2,(3+sqrt(3))/2) is a. 0 b. sqrt(2) c. 3+sqrt(3) d. none of these

On the line segment joining (1, 0) and (3, 0) , an equilateral triangle is drawn having its vertex in the fourth quadrant. Then the radical center of the circles described on its sides. (a) (3,-1/(sqrt(3))) (b) (3,-sqrt(3)) (c) (2,-1sqrt(3)) (d) (2,-sqrt(3))

AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60o . He moves away from the pole along the line BC to a point D such that C D""=""7""m . From D the angle of elevation of the point A is 45o . Then the height of the pole is (1) (7sqrt(3))/2 1/(sqrt(3)-1)m (2) (7sqrt(3))/2 sqrt(3)+1m (3) (7sqrt(3))/2 sqrt(3)-1m (4) (7sqrt(3))/2 sqrt(3)+1m

If the angles of a triangle are in the ratio 4:1:1, then the ratio of the longest side to the perimeter is (a) sqrt(3):(2+sqrt(3)) (b) 1:6 (c) 1:2+sqrt(3) (d) 2:3