Some Applications Of Trigonometry

In this figure; AC is the rope which is 20 m long and we have to find AB (height of the pole).

In Δ ABC;

where p = perpendicular and h = hypotenuse

Hence, Height of the pole, AB = 10 m

Let BAC be the tree.

Now, due to storm AC is broken and leans to make a shape as right-angled triangle shown in figure.

AC is the broken part of the tree and AB is the part which is still upright.

Hence,

Distance from tree foot to where top touches, BC = 8 m

And,

∠ BCA = 30^o

Clearly, Sum of AB and AC will give the height of the tree.

In Δ ABC:

AC =

Hence,

AB =

Now,

Height of tree= AB + AC

=

=

= 8√3 m

Hence, height of the tree is 8√3 m

In the first case:

Height of slide= 1.5 m and angle of elevation = 30°

Now,

Hence, h = 3 m

In the second case:

Height of slide, = 3 m, angle of elevation = 60°

Hence, h = 2 m

Therefore, the length of the slide in the first case and the second case are 3 m and 2respectively.

In this question:

b = 30 m,

Angle of elevation = 30°

And

p = height of tower = ?

[ p = perpendicular, b = base]

Rationalizing the value of p, we get







Hence the height of the tower is 10√3 m

To find: Length of String, AC

Given:

Height of kite from the ground = BC = 60 m, angle of elevation = 60° and length of string = AC = ?

Now we know that,



So in the given triangle,

Hence, length of the string is 40√3 m

Let AB be the boy, and CD be the building and angles becomes 60°, as the boy reaches E.

The angle of elevation in first case = 30°

And

The angle of elevation, in the second case = 60°.

Difference between bases in the first case and second case will give the distance covered by the boy.

1st case:
here p = height of the building - the height of the boy
= 30 - 1.5
= 28.5 m

b = 28.5√3

2^nd case:

Hence, the distance covered by the boy:

= 28.5

= 19√3 m

Since the building is vertical.
∠QPO = 90°

_{In ∆OPQ,}

tan 45° =

1 =

OP = 20 -------(1)

Now in ∆OPR

tan 60° =

=

=

20√3 = h+20

h = 20√3-20

h = 20(√3-1) m.

Therefore the height of transmission tower is 20(√3-1) m.

Let AD be the height of the statue which is 1.6 m,

And DB be the height of the pedestal.

And C be the point where the observer is present.

To find: height of pedestal, BD
Now,

In Δ ABC,

where p = perpendicular and b = base

In Δ DBC,

On comparing (i) and (ii),

Hence, height of pedestal is 0.8(√3 + 1)m

Let us take AB as building and
Let us take DC as tower = 50 m,

∠ACB = 30°

And,

∠ DBC = 60°;

AB= ?

In Δ DCB;

In Δ ACB

AB= 16.67 m

Let AB and DE be the two poles, and C be the point of observation.
Given,
width of road, BD = 80 m
Angle of elevation to AB, ∠ACB = 30°
Angle of elevation to DE, ∠ ECD = 60°
To find: Height of buildings AB and DE.

In Δ ACB

In Δ EDC:

We know that AB = ED as the poles are of same height.

Hence, from (i) and (ii),

cross multiplying, we get

Now, using the value of BC in (i),

AB = 20√3 m
Now from triangle EDC,

Therefore,
Height of Pole = 20√3 m
Distances of poles from observing point = 60 m and 20 m

According to this figure:

CD = 20 m,

∠ACB = 60°,

∠ ADB = 30°,

AB = ? and BD = ?

In Δ ABC,

In Δ ABD,

𝐹rom (i) and (ii),

Using the value of BC in (i) we get;

AB = 10√3 m which is the height of the tower

Width of canal = 10 m

Let us take AB as a building, with AB = 7 m
and CE as a cable tower
Angle of elevation from top of tree to top of cable tower∠EAD = 60°,
and angle of depression from the top of the building to the bottom of the tower is, ∠CAD = 45°
Also,
∠CAD = ∠ACB [Alternate angles]

Clearly,
AB = DC and EC = ED + DC ?

In Δ ABC:

where p = perpendicular and b = base, therefore

AB = BC = 7 m

In Δ EDA,

However,

The height of tower can be calculated as:

EC = ED + DC

= 7√3 + 7

= 7(√3 +1) m

A and D are the two ships and the distance between the ships is AD. BC is the lighthouse and is the height of the lighthouse. observer is at point C
Give: BC = height of lighthouse = 75m

∠CAB = 45°,

∠CDB = 30°

To Find: DA

In Δ ABC,

[ p = perpendicular and b = base of the right angled triangle]

AB=75 m [ tan 45º = 1]

In Δ CDB,

AD + AB= 75√3 m

DA = 75( √3 - 1) m
Hence the distance between the ships is 75( √3 - 1) m

From this figure:

BF = height of girl = 1.2 m,

AG = height of balloon from ground initially = 88.2 m,

AC = AG – GC

= 88.2 – 1.2

= 87 m

∠EBD = 30° and ∠ABC = 60°,

CD = ?

In Δ ABC;

In Δ ABC,

Now, the distance covered by the balloon will be:

CD = BD – BC

CD = 58√3 m

The diagram is:

Let AB is the tower and AD is the highway.

Now from triangle ADB,

tan 30° = AB/AD

⇒ 1/√3 = AB/AD

⇒ AB = AD/√3 .............(1)

Again from triangle ACB

tan 60° = AB/AC

⇒ √3 = AB/AC

⇒ AB = AC√3 ........(2)

from equation 1 and 2

AD/√3 = AC√3

⇒ (DC + CA)/√3 = AC√3

⇒ DC + CA = AC√3×√3

⇒ DC + CA = 3AC

⇒ 3AC - Ac = CD

⇒ 2AC = CD

⇒ AC = CD/2

Since time taken by car to cover CD = 6 Second

So time taken by car to cover AC = 6/2 = 3 seconds.

In this figure

Let ∠ADB = θ and ∠ACB = 90°-θ [ As angles are complementary, their sum will be equal to 90°, and if one is θ other will be 90°-θ]
In Δ ABD;

In Δ ABC,

But we know that,

And also by complementary angle formula that tan(90°-θ) = cot θ
tan(θ) tan(90°-θ) = 1

Therefore,

AB = 6 m

Hence proved.