Question 28 - CBSE Sample Paper Class 10 - 2017-18

Last updated at May 29, 2018 by Teachoo

The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Question 28
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30 and 60 , respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
Let building be AB & tower be CD
Height of building = AB = 50 m
Let, Height of tower = PC
And, distance between tower and building = AC
Angle of depression to top of building = QDB = 30
Angle of depression to bottom of building = QDA = 60
We have to find
height of tower i.e. CD
And distance between tower &building i.e. AC
Draw BE parallel to AC & DQ
Lines DQ & BE are parallel,
And BD is the transversal
DBE = QDB
DBE = 30
Now,
AC and BE are parallel lines.
So, AC = BE
Lines DQ & AC are parallel,
And AD is the transversal
DAC = QDA
DAC = 45
Similarly, AB and CE are also parallel lines.
So, CD = AB
CD = 50 m
Also,
Since DC is perpendicular AC
DEB = DCA = 90
In right angle triangle DBE
tan B = ( )/(Side " " )
tan B = (" " )/
tan 30 = /
1/ 3 = (" " )/
BE = DE 3
In right angle triangle PAC
tan A = ( " " )/( " " )
tan 60 = (" " )/
3 = (" " )/
AC = (" " )/ 3
"BE = " (" " )/ 3
From (1) & (2)
DE 3 = (" " )/ 3
DE 3 3 = DC
3DE = DC
3DE = DE + EC
3DE DE = EC
2DE = EC
2DE = 50
DE = 50/2
DE = 25 m
Height of tower = DC = DE + EC = 25 + 50 = 75 m
From (1)
BE = DE 3
Putting DE = 25
BE = 25 3 m
So, AC = BE = 25 3 m
Distance between two buildings = AC = 25 m

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