Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Some Applications of Trigonometry

IN the figure

angle made by the rope with the ground level is 30° and Length of the rope is 20 m and we have to
determine the height of the pole AB=?

So,.

AC = 20 m and angle θ= 30°

In right ΔABC,

sin 30° = AB/AC

^{ 1/2 = AB/20}

^{ AB = 20/2}

^{AB = 10}

Therefore, the height of the pole is 10 m.

Let's the original height of tree be AC due to storm tree broke from point .the broken part of tree is BC touch at ground at D making a angle 30°

BC = 8 m

From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right ΔABC,

cos30° = BC/AC

Here, cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

Also,

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.

As per contractor’s plan,

Let AC is the slide for the children below the age of 5 yr

So , BC=1.5 ∠CAB=30°

In right ΔABC,

sin 30° = AB/AC

1/2 = 1.5/AC

AC = 3

Let PR is the slide for the children below the age of elder child

So , PQ=3m ∠QPR=60°

Also,

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3

Hence, length of the slide for below 5 = 3 m and

Length of the slide for elders children = 2√3 m

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

here ∠ACB=30°

In right ABC

tan 30° = AB/BC

1/√3 = AB/30

⇒ AB = 10√3

Thus, the height of the tower is 10√3 m.

Let BC is Height of the kite from the ground, BC = 60 m

AC is length of the string from the ground which make an angle ∠ACB=60°

From the above figure,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

Thus, the length of the string from the ground is 40√3 m.

Let the boy initially stand at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

From figure,

XY = CD.

Height of the building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

Measure of AB is 28.5 m

In right ΔABD,

tan 30° = AB/BD

1/√3 = 28.5/BD

BD = 28.5√3 m

Again,

In right ΔABC,

tan 60° = AB/BC

√3 = 28.5/BC

BC = 28.5/√3 = 28.5√3/3

Therefore, the length of BC is 28.5√3/3 m.

XY = CD = BD – BC = (28.5√3-28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.

Thus, the distance boy walked towards the building is 57/√3 m.

transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Let BC be the 20 m high building.

D is the point on the ground from where the elevation is taken.

Height of transmission tower = AB = AC – BC

From figure, In right ΔBCD,

tan 45° = BC/CD

1 = 20/CD

CD = 20

Again,

In right ΔACD,

tan 60° = AC/CD

√3 = AC/20

AC = 20√3

Now, AB = AC – BC = (20√3-20) = 20(√3-1)

Height of transmission tower = 20(√3 – 1) m.

Let AB be the height of statue.

D is the point on the ground from where the elevation is taken.

To Find: Height of pedestal = BC = AC-AB

From figure,

In right triangle BCD,

tan 45° = BC/CD

1 = BC/CD

BC = CD …..(1)

Again,

In right ΔACD,

tan 60° = AC/AD

√3 = ( AB+BC)/CD

√3CD = 1.6 + BC

√3BC = 1.6 + BC (using equation (1)

√3BC – BC = 1.6

BC(√3-1) = 1.6

BC = 1.6/(√3-1) m

BC = 0.8(√3+1)

Thus, the height of the pedestal is 0.8(√3+1) m.

Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building respectively.

In right ΔBCD,

tan 60° = CD/BC

√3 = 50/BC

BC = 50/√3 …(1)

Again,

In right ΔABC,

tan 30° = AB/BC

1/√3 = AB/BC

Use result obtained in equation (1)

AB = 50/3

Thus, the height of the building is 50/3 m.

Let AB and CD be the poles of equal height.

O is the point between them from where the height of elevation taken. BD is the distance between the poles.

As per above figure, AB = CD,

OB + OD = 80 m

Now,

In right ΔCDO,

tan 30° = CD/OD

1/√3 = CD/OD

CD = OD/√3 … (1)

Again,

In right ΔABO,

tan 60° = AB/OB

√3 = AB/(80-OD)

AB = √3(80-OD)

AB = CD (Given)

√3(80-OD) = OD/√3 (Using equation (1))

3(80-OD) = OD

240 – 3 OD = OD

4 OD = 240

OD = 60

Putting the value of OD in equation (1)

CD = OD/√3

CD = 60/√3

CD = 20√3 m

Also,

OB + OD = 80 m

OB = (80-60) m = 20 m

Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and

60 m respectively.

DC = 20 m (given)

As per given diagram, In right ΔABD,

tan 30° = AB/BD

1/√3 = AB/(20+BC)

AB = (20+BC)/√3 … (i)

Again,

In right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

AB = √3 BC … (ii)

From equation (i) and (ii)

√3 BC = (20+BC)/√3

3 BC = 20 + BC

2 BC = 20

BC = 10

Putting the value of BC in equation (ii)

AB = 10√3

This implies, the height of the tower is 10√3 m and the width of the canal is 10 m.

Let AB be the building of height 7 m and EC be the height of the tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°.

EC = DE + CD

Also, CD = AB = 7 m. and BC = AD

To Find: EC = Height of the tower

Design a figure based on given instructions:

In right ΔABC,

tan 45° = AB/BC

1= 7/BC

BC = 7

Since BC = AD

So AD = 7

Again, from right triangle ADE,

tan 60° = DE/AD

√3 = DE/7

⇒ DE = 7√3 m

Now: EC = DE + CD

= (7√3 + 7) = 7(√3+1)

Therefore, Height of the tower is 7(√3+1) m. Answer!

Let AB be the lighthouse of height 75 m. Let C and D be the positions of the ships.

30° and 45° are the angles of depression from the lighthouse.

Draw a figure based on given instructions:

To Find: CD = distance between two ships

Step 1: From right triangle ABC,

tan 45° = AB/BC

1= 75/BC

BC = 75 m

Step 2: Form right triangle ABD,

tan 30° = AB/BD

1/√3 = 75/BD

BD = 75√3

Step 3: To find measure of CD, use results obtained in step 1 and step 2.

CD = BD – BC = (75√3 – 75) = 75(√3-1)

The distance between the two ships is 75(√3-1) m. Answer!

Let the initial position of the balloon be A and final position be B.

Height of balloon above the girl height = 88.2 m – 1.2 m = 87 m.

To Find: Distance travelled by the balloon = DE = CE – CD

Let us redesign the given figure as per our convenient

tan 30° = BE/CE

1/√3= 87/CE

CE = 87√3

Step 2:

In right ΔADC,

tan 60° = AD/CD

√3= 87/CD

CD = 87/√3 = 29√3

Step 3:

DE = CE – CD = (87√3 – 29√3) = 29√3(3 – 1) = 58√3

Distance travelled by the balloon = 58√3 m.

Let AB be the tower.

D is the initial and C is the final position of the car respectively.

Since man is standing at the top of the tower so, Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

Step 1: In right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

BC = AB/√3

AB = √3 BC

Step 2:

In right ΔABD,

tan 30° = AB/BD

1/√3 = AB/BD

AB = BD/√3

Step 3: Form step 1 and Step 2, we have

√3 BC = BD/√3 (Since LHS are same, so RHS are also same)

3 BC = BD

3 BC = BC + CD

2BC = CD

or BC = CD/2

Here, distance of BC is half of CD. Thus, the time taken is also half.

Time taken by car to travel distance CD = 6 sec. Time taken by car to travel BC = 6/2 = 3 sec.

Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As per question,

In right ΔABC,

tan x = AB/BC

tan x = AB/4

AB = 4 tan x … (i)

Again, from right ΔABD,

tan (90°-x) = AB/BD

cot x = AB/9

AB = 9 cot x … (ii)

Multiplying equation (i) and (ii)

AB^{2} = 9 cot x × 4 tan x

⇒ AB^{2} = 36 (because cot x = 1/tan x

⇒ AB = ± 6

Since height cannot be negative. Therefore, the height of the tower is 6 m.

Hence Proved.