**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