TOPICS

Unit-9(Examples)

Some Applications of Trigonometry

**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.

Let 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. Now, tan 60° = AB/BC √3 = AB/15 AB = 15√3 Hence, the height of the tower is 15√3 m.

**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 Figure). 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)

Here 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. BD/BC = sin 60° 3.7/BC = √3/2 BC = 7.4/√3 = 4.28 m(Approx) i.e., Length of ladder = 4.28 m Now, DC/BD = cot 60° DC/3.7 = 1/√3 DC = 3.7/√3 = 2.14 m(Approx) Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

**Example-3 :-** 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?

Here, AB is the chimney, CD the observer and ∠ ADE the angle of elevation. 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 tan 45 = AE/DE 1 = AE/28.5 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)

AB denotes the height of the building, BD the flagstaff and P the given point. Height of building = 10 m Since, we know the height of the building AB, we will first consider the right Δ PAB. tan 30° = AB/AP 1/√3 = 10/AP AP = 10√3 m 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 Δ PAD tan 45° = AD/AP = (10 + x)/10√3 1 = (10 + x)/10√3 10√3 = 10 + x x = 10√3 - 10 x = 10(√3 - 1) x = 7.32 m 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.

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 two right triangles ABC and ABD In Δ ABC, tan 60° = AB/BC √3 = h/x h = √3x ..........(1) In Δ ABD, tan 30° = AB/BD 1/√3 = h/(x + 40) ..........(2) Put the value of (1) in to (2) 1/√3 = √3x/(x + 40) x + 40 = 3x 3x - x = 40 2x = 40 x = 40/2 x = 20m Put the value of x = 20 m in to (1) h = 20√3 m 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.

PC denotes the multistoryed 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. Observe that PB is a transversal to the parallel lines PQ and BD. Therefore, ∠ QPB and ∠ PBD are alternate angles, and so are equal. So ∠ PBD = 30°. Similarly, ∠ PAC = 45°. In right Δ PBD, tan 30 = PD/BD 1/√3 = PD/BD BD = √3 PD ....(1) In right Δ PAC, tan 45 = PC/AC 1 = PC/AC PC = AC Also, PC = PD + CD, therefore, PD + CD = AC Since, AC = BD and CD = AB = 8m, we get PD + 8 = BD = PD√3 PD = 8/(√3 - 1) By rationalising of denomenator PD = 8(√3 + 1)/[(√3 - 1)(√3 - 1)] PD = 8(√3 + 1)/(3 - 1) PD = 8(√3 + 1)/2 PD = 4(√3 + 1) m 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.

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. Now, AB = AD + DB In right Δ APD, ∠ A = 30°. So, tan 30 = PD/AD 1/√3 = 3/AD AD = 3√3 m Also, in right Δ 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

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