Home
Class 12
MATHS
Let a vertical tower A b have its end A ...

Let a vertical tower `A b` have its end `A` on the level ground. Let `C` be the mid point of `A B` and `P` be a point on the ground such that `A P=2A Bdot` If `/_B P C=beta,` then `tanbeta` is equal to : `2/9` (2) `4/9` (3) `6/7` (4) `1/4`

Promotional Banner

Topper's Solved these Questions

  • THREE DIMENSIONAL GEOMETRY

    JEE MAINS PREVIOUS YEAR|Exercise All Questions|22 Videos

  • VECTOR ALGEBRA

    JEE MAINS PREVIOUS YEAR|Exercise All Questions|19 Videos

Similar Questions

Explore conceptually related problems

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If sqrtBPC = beta , then tan beta is equal to

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP-2AB. If BPC-?, then tan ? is equal to 12. 6 (2) 4 2 4

AB is a vertical pole. The end A is on the level ground .C is the middle point of AB. P is a point on the level ground . The portion BC subtends an angles beta at P. If AP = nAB, then tan beta =

AB is a vertical pole and C is its mid point. The end A is on the level ground and P is any point on the level ground other than A the portion CB substends an angle beta at P. If AP:AB=2:1, then beta =

AB is a vertical pole resting at the end A on the level ground. P is a point on the level ground such that AP = 3AB and C Is the midpoint of AB. If AC and CB subtend angles alpha and beta , respectively, at P, then the value of tan beta is

Let P and Q be points on the line joining A(-2, 5) and B(3, 1) such that AP = PQ = QB. If mid-point of PQ is (a, b), then the value of (b)/(a) is

Let A be a fixed point (0,6) and B be a moving point (2t,0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :

A(4,2), B(6,5) and C(1,4) are the vertices of triangle ABC Find coordinates of points P on AD such that AP : PD= 2: 1 .