NCERT Solutions Class 10th Maths Chapter – 9 Applications of Trigonometry Examples

Example 1. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Solution: First let us draw a simple diagram to represent the problem (see Fig. 9.4). Here AB represents the tower, CB is the distance of the point from the tower and ACB is the angle of elevation. We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angled at B. To solve the problem, we choose the trigonometric ratio tan 60° (or cot 60°), as the ratio involves AB and BC.Now, tan 60° = AB/BC

i,e., √3 = AB/15
i,e., AB = 15√3

Example 2. An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take √3 = 1.73)

Solution: In Fig. 9.5, the electrician is required to
reach the point B on the pole AD.
So, BD = AD – AB = (5 – 1.3)m = 3.7 m.
Here, BC represents the ladder. We need to find its length, i.e., the hypotenuse of the right triangle BDC.
Now, can you think which trigonometric ratio should we consider?
It should be sin 60°.
So, BD/BC = sin 60° or 3.7/BC = √3/2
Therefore, BC = 3.7 × 2/√3 = 4.28 m (approx.)
i.e., the length of the ladder should be 4.28 m
Now, DC/BD= cot 60° = 1/√3
i.e., DC = 3.7/√3 = 2.14 m (approx.)
Therefor, she should place the foot of the ladder at a distance of 2.14 m from the pole.

Example 3. An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?

Solution : Here, AB is the chimney, CD the observer and ∠ADE the angle of elevation (see Fig. 9.6). In this case, ADE is a triangle, right-angled at E and we are required to find the height of the chimney.
We have
AB = AE + BE = AE + 1.5
and
DE = CB = 28.5 m
To determine AE, we choose a trigonometric ratio, which involves both AE and
DE. Let us choose the tangent of the angle of elevation.
Now, tan 45° = AE/DE
i.e., 1 = AE/28.5
therefore, AE = 28.5
So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m.

Example 4. From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take √3 = 1.732)

Solution: In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e., DB and the distance of the building from the point P, i.e., PA.
Since, we know the height of the building AB, we will first consider the right △ PAB.
We have tan 30° = AB/AP
i.e., 1/c = 10/AP
Therefore, AP = 10√3
i.e., the distance of the building from P is 10√3 m = 17.32 m.
Next, let us suppose DB = x m. Then AD = (10 + x) m.
Now, in right △ PAD, tan 45° = AD/AP = 10 + x/10√3
Therefore, 1 = 10 + x/10√3
i.e., x = 10 (√3 – 1) = 7.32
So, the length of the flagstaff is 7.32 m.

Example 5. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Solution: In Fig. 9.8, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer than BC.So, DB = (40 + x) m
Now, we have tow right triangles ABC and ABD.
In △ABC, tan 60° = AB/BC
or, √3 = h/x …(1)
In △ABD, tan 30° = AB/BD
i.e., 1/√3 = h/x + 40 …(2)
From (1), we have h = x√3
Putting this value in (2) we get (x√3)√3 = x + 40, i.e., 3x = x + 40
i.e., x= 20
So, h = 20√3 [from (1)]
therefore, the height of the tower is 20√3 m.

Example 6. 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 the multi-storeyed building and the distance between the two buildings.

Solution: In Fig. 9.9, PC denotes the multi-storyed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC and the distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. Therefore, ∠QPB and ∠PBD are alternate angles, and So are alternate are equal. So PBD = 30°. Similarly, ∠PAC = 45°. In right ∠PBD, we have
PD/BD = tan 30° = 1/√3 or BD = PD √3
In right △PAC, we have
PC/AC = tan 45° = 1
PC = AC
PC = PD + DC, therefore, PD + DC = AC.
Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD√3 (Why?)
So, the height of the multi-storeyed building is {4(√3 + 1) + 8}m = 4(3 + √3) m and the distance between the two buildings is also 4(3 + √3) m

Example 7. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Solution: In Fig 9.10, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e., DP = 3 m. We are interested to determine the width of the river, which is the length of the side AB of the D APB.
Now, AB = AD + DB
In right △ APD, ∠ A = 30°.
So, tan 30° = PD/AD
i.e., 1/√3 = 3/AD or AD = 3√3 m
Also, in right △ PBD, B = 45°. So, BD = PD = 3 m.
Now, AB = BD + AD = 3 + 3√3 = 3 (1 + √3 ) m.
Therefore, the width of the river is 3(√3+1)m.