Question

Two points A and B are on the same side of a tower in the same straight line with its base. The angle of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15m, then find the distance
between these points.

Quality answer needed! no Spamming !​

Answer 1

Refer to the attachment for the diagram!!

Given:-

angle of depression for point A = 60⁰

angle of depression for Point B = 45⁰

Height of the tower = 15 m

Concept:-

angle of depression of Point A = angle of elevation by point A = 60⁰

angle of depression of Point B = angle of elevation by point B = 45⁰

To find:-

distance between these two points i.e AB

Formula to be used:-

$\underline{\boxed{\bf TanØ = \dfrac{Opposite\:side}{Adjacent\:side}}}$

Solution:-

let us assume AC as x

and AB as y

$\implies$ $\bf Tan60⁰ = \dfrac{DC}{AC}$

$\implies$ $\bf Tan60⁰ = \dfrac{15}{x}$

$\implies$ $\bf √3 = \dfrac{15}{x}$

$\implies$ $\bf √3x = 15$

$\implies$ $\bf x = \dfrac{15}{√3}$ = AC ... $\bf eq_{1}$

$\implies$ $\bf Tan45⁰ = \dfrac{AC}{BC}$

$\implies$ $\bf Tan45⁰ = \dfrac{15}{y+x}$

$\implies$ $\bf 1 = \dfrac{15}{x+y}$

$\implies$ $\bf x+y = 15$

$\implies$ $\bf \dfrac{15}{√3}+y = 15$ ... from $\bf eq_{1}$

$\implies$ $\bf y = 15-\dfrac{15}{√3}$

$\implies$ $\bf y = \dfrac{15√3-15}{√3}$

$\implies$ $\bf y = \dfrac{15(√3-1)}{√3}$

$\implies$ $\bf y = 5√3(√3-1) m$

hence, AB = $\bf y = 5√3(√3-1) m$