sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
The angles of depression of the top and the bottom of an 8 m tall building
from the top of a multi-storeyed building are 30 and 45 respectively.
Find the height of tire multi-storeyed building and the distance between the two buildings,
correct to two decimal places.
question no 32 , heights and distances , ICSE board, ML Aggarwal
Consider AB as the height and CD as the building
The angles of depression from A to C and D are 300 and 450
∠ACE = 300 and ∠ADB = 450
CD = 8 m
Take AB = h and BD = x
From the point C
Construct CE parallel to DB
CE = DB = x
EB = CD = 8 m
AR = AB – EB = h – 8
In right triangle ADB
tan θ = AB/DB
Substituting the values
tan 450 = h/x
So we get
1 = h/x
x = h
In right triangle ACE
tan 300 = AE/CE
Substituting the values
1/√3 = (h – 8)/ h
By further calculation
h = √3h – 8√3
So we get
√3h – h = 8√3
h (√3 – 1) = 8√3
h = 8√3/(√3 – 1)
Multiply and divide by √3 + 1
h = 8√3/ (√3 – 1) × (√3 + 1)/ (√3 + 1)
h = 8 (3 + √3)/ (3 – 1)
Here
h = 8 (3 + 1.732)/ 2
h = 4 × 4.732
h = 18.928
h = 18.93 m
x = h = 18.93 m
Here
Height of multi-storeyed building = 18.93 m
Distance between the two buildings = 18.93 m