If $$BD, BE$$ and $$CF$$ are the medians of a $$\vartriangle ABC$$ then $$\left( {A{D^2} + B{E^2} + C{F^2}} \right):\left( {B{C^2} + C{A^2} + A{B^2}} \right)$$ is equal to
From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression of a boat is 15°. The distance of the boat from the foot of the light house is
A.
$$\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)60\,{\text{metres}}$$
B.
$$\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)60\,{\text{metres}}$$
C.
$${\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}{\text{metres}}$$
In a triangle $$ABC,$$ angle $$A$$ is greater than angle $$B.$$ If the measures of angles $$A$$ and $$B$$ satisfy the equation $$3\sin x - 4{\sin ^3}x - k = 0, 0 < k < 1,$$ then the measure of angle $$C$$ is
In a triangle $$ABC,$$ $$\angle B = \frac{\pi }{3}{\text{ and }}\angle C = \frac{\pi }{4}.$$ Let $$D$$ divide $$BC$$ internally in the ratio 1 : 3 then $$\frac{{\sin \angle BAD}}{{\sin \angle CAD}}$$ is equal to