JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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Application +4 | -1

A stone is projected vertically upward moves such that its distance feet in time t sec is given by

s=112 t-16 t^2

then the maximum height reached by the stone in (feet) is

View answer and explanation

Correct option: A

196

Option A (196) is correct because it is the maximum height reached by the stone, calculated by finding the time when the velocity is zero and substituting it into the distance equation.

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{p_1},{p_2},{p_3}

are altitudes of a

\Delta \;ABC

drawn from the vertices

A,B,C

and

\Delta

the area of the triangle, then

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

is equal to If the sides of the triangles are

5k,\;6k,\;5k

and radius of incircle is 6, the value of

k

is equal to In a

\Delta \,ABC

, if

a = \left( {b - c} \right)\sec \theta

, then

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

\Delta \;ABC,\frac{{b + c}}{a}

is equal to If

\Delta = {a^2} - {\left( {b - c} \right)^2}

, where

\Delta

is the area of

\Delta \;ABC

, then

\tan A

is equal to If

A

is the area and

2 s

is the sum of the sides of a triangle, then