A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 60º at the foot of the tower, and the angle of elevation of the top of the tower from A or B is 30º. The height of the tower is
In a triangle PQR, ∠R =π/2. If (P/2) and tan (Q/2) are the roots of ax2 +bx+ c = 0, a ≠ 0 then
a = b + c
c = a + b
b = c
b = c
In a triangle, ABC, let ∠C = π/2 . If r is the in radius and R is the circumradius of the triangle ABC, then 2 (r + R) equals
b + c
a+b
a + b + c
a + b + c
Let α, β be such that π < α - β < 3π. If sinα + sinβ = -21/65 and cosα + cosβ = -27/65, then the value of cos α-β/2 is
6/65
6/65
If then the difference between the maximum and minimum values of 2 u is given by
2(a2 + b2)
2(a2-b2)
(a+b)2
(a+b)2
The sides of a triangle are sinα, cosα and for some 0 < α < π/2 . Then the greatest angle of the triangle is
60o
120o
360o
360o
B.
120o
Greatest side is by applying cos rule we get greatest angle = 120ο