Question

the angle of elevation of the top of a building form the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is the angle of elevation of a jet plane from a point A on the ground is60°
​

Answer 1

Answer:

The situation in the problem is shown in the attached figure.

Given:

CE=48 m

∠EBC=60  o

 ∠ACB=30  o

 To find: AB

Using trigonometric identity,

tan(∠EBC)=  

BC

EC -(i)

tan(∠ACB)=  

BC

AB  -(ii)

Dividing (ii) by (i),

EC

AB

​ =  

tan(∠EBC)

tan(∠ACB)

​Substituting values,

48

AB

​ =  

tan60  o

tan30  o

 48

AB =  3

​1/  3

AB=  3

48

​= 16 m

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Answer 2

Answer:

Given height of tower CD=50m

Let the height of the building, AB=h

In right angled △BDC,

$tan \: 60° = \frac{cd}{bd} </p><p>$

$\sqrt{3} \: = \frac{50}{bd}$

$bd \: = \frac{50}{\sqrt{3} } \: m$

In right angled △ABD,

$tan \: 30° \: = \frac{ab}{bd}$

$\frac{1}{ \sqrt{3 } } = \frac{h}{50} \sqrt{3}$

$\therefore \: \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3h} }{50}$

$\therefore \: h = \frac{50}{3} = 16.66 \: m$

$\therefore \: the \: hight \: of \: the \: building \: is \: 16.66 \: m$

$\mathfrak \: HOPE \: THIS \: HELPS \: YOU \:$