Trigonometry Exercise 13.5 – Questions:
- Find the value of ‘x’:
(i)
(ii)
(iii)
(iv)
(v)
II.
- A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30˚. Find the height of the tower.
- From the top o a building 50√3 m high, the angle of depression of an object on the ground is observed to be 45˚. Find the distance of the object from the building.
- A tree is broken over by wind forms a right angled triangle with the ground. If the broken part makes an angle of 60˚ with the ground and the top of the tree is now 20m from its base, how tall was the tree?
- The angle of elevation of the top of a flag-post from a point on a horizontal ground is found to be 30˚. On walking 6 m towards the post, the elevation increased by 15˚. Find the height of the flag-post.
- The angles of elevation of the top of the cliff as seen from the top and bottom of a building are 45˚ and 60˚ respectively. If the height of the building is 24 m, find the height of the cliff.
- From the top a building 16 m high, the angular elevation of the top of a hill is 60˚ and the angular depression of the foot of the hill is 30˚. Find the height of the hill.
- Find the angle of depression if an observer 150 m tall looks at the tip of his shadow which is 150√3 cm from his foot.
- From a point 50 m above the ground the angle of elevation of a cloud is 30˚ and the angle of depression of its reflection in water is 60˚. Find the height of the cloud above the ground.
Trigonometry Exercise 13.5 – Solutions:
- Find the value of ‘x’:
(i)
Solution:
In ∆ABC , ∠ACB = 45˚
tanθ = AB/BC
tan 45˚ = x/60
1 = x/60
x = 60 m
(ii)
Solution:
In ∆PRQ , ∠RQP = 60˚
tanθ = RP/PQ
tan 60˚ = 90/x
√3 = 90/x
√3 x = 90
x = 30√3 units
(iii)
Solution:
In ∆LKM , ∠KLM = 30˚
tanθ = KM/LM
tan 30˚ = x/100
1/√3 = x/100
x√3 = 100
x = 100/√3 units
(iv)
Solution:
In ∆XYZ , ∠XZY = 45˚
cosθ = YZ/XZ
cos 45˚ = x/100
1/√2 = x/100
x√2 = 100
x = 100/√2 units
(v)
Solution:
In ∆DEF , ∠DEF = x
tanθ = DF/FE
tan x = 75/75
tan x = 1
We know, tan 45˚ = 1
tan x = tan 45˚
x = 45˚
II.
- A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30˚. Find the height of the tower.
Solution:
In ∆ABC , ∠ACB = 30˚
tanθ = AB/BC
tan 30˚ = x/300
1/√3 = x/300
x√3 = 300
x = 300/√3 = 100√3
- From the top of a building 50√3 m high, the angle of depression of an object on the ground is observed to be 45˚. Find the distance of the object from the building.
Solution:
Let PQ be the height f the building, PQ = 50√3, QR be the distance between the building and object.
Angle of depression = 45˚
Since, PM||QR, ∠MPR = ∠PRQ (alternate angles)
∠MPR = 45˚
Therefore, ∠PRQ = 45˚
In ∆PQR , ∠MPR = 45˚
tanθ = PQ/QR
tan 45˚ = 50√3/QR
1 = 50√3/QR
QR = 50√3 m
- A tree is broken over by wind forms a right angled triangle with the ground. If the broken part makes an angle of 60˚ with the ground and the top of the tree is now 20m from its base, how tall was the tree?
Solution:
Let AC is the full length of the tree. AB is broken and bends as BD. Therefore AB = BD.
Given, CD = 20 m and ∠BDC = 60˚
In ∆BCD , ∠BDC = 60˚
tanθ = BC/CD
tan 60˚ = 20/CD
√3 = BC/20
20√3 = BC
cosθ = CD/BD
1/2 = 20/BD
40 = BD
To find the full length of the tree , AC = BC + BD(since AB = BD)
= 20√3 + 40
= 20(√3 + 2) m
- The angle of elevation of the top of a flag-post from a point on a horizontal ground is found to be 30˚. On walking 6 m towards the post, the elevation increased by 15˚. Find the height of the flag-post.
Solution:
Given ∠BAC = 30˚, ∠BDC = 45˚
tan30˚ = BC /AD + DC
1/√3 = BC/6+x
√3 .BC = 6 + x
BC = 6+x/√3 ………….(1)
tan45˚ = BC/DC
1 = BC/x
BC = x…………..(2)
Substitute (2) in (1),
x = 6 + x/√3
√3.x = 6 + x
6 = √3.x – x
6 = x(√3 – 1)
x = 6/(√3 – 1)
- The angles of elevation of the top of the cliff as seen from the top and bottom of a building are 45˚ and 60˚ respectively. If the height of the building is 24 m, find the height of the cliff.
Solution:
Height od the building i.e., AD = BC = 24 m. Similarly, AB = DC,
In ∆MCD , ∠MDC = 60˚
tan θ = MC/DC
tan60˚ = 24+x/DC
√3 = 24+x/DC
DC.√3 = 24 + x
DC = 24+x/√3
In ∆AMB , ∠MAB = 45˚, MB = x
tan θ = MB/AB
tan45˚ = x/(24+x/√3)
1 = x√3/24+x
√3x = 24 + x
√3x – x = 24
(√3 – 1)x = 24
x = 24/√3 – 1
Height of the cliff, MC = MB + BC = (24 + 24/√3 – 1) m
- From the top a building 16 m high, the angular elevation of the top of a hill is 60˚ and the angular depression of the foot of the hill is 30˚. Find the height of the hill.
Solution:
In ∆DCE , ∠CED = 30˚, DE = x
tan θ = CD/DE
tan30˚ = 16/x
1/√3 = 16/x
x = 16√3
In ∆ABC , ∠BCA = 60˚, BC = DE = x
tan θ = AB/BC
tan 60˚ = AB/x
√3 = AB/16√3
16×3 = AB
48 = AB
Therefore, the height of the hill = AB + BE = 48 + 16 = 64 m
- Find the angle of depression if an observer 150 m tall looks at the tip of his shadow which is 150√3 cm from his foot.
Solution:
In ∆ABC , AB = 150 cm and BC = 150√3
tan θ = AB/BC = 150/150√3
tan θ = 1/√3
We know, tan30˚ = 1/√3
Therefore, tan θ = tan30˚
Hence, θ = 30˚
- From a point 50 m above the ground the angle of elevation of a cloud is 30˚ and the angle of depression of its reflection in water is 60˚. Find the height of the cloud above the ground.
Solution:
Let AC = h
BE = CD = 50 m
A’D = AD = 50 + h
In ∆ABC , ∠ABC = 60˚,
tan θ = AC/BC = h/BC
tan 30˚ = h/BC
1/√3 = h/BC
√3.h = BC
Now, in ∆A’CB ,
tan60˚ = CA’/BC = CD+DA’/BC
√3 = 50+50+h/√3h = 100+h/√3h
3h = 100 + h
2h = 100
h = 50 m
Thus, height of the cloud from the surface of lake = AD+A’D = 50 + 50 = 100m