Some Applications of Trigonometry

NCERT Exercise 9.1 solution Q 13 to 14

Question: (13) As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are 30^o and 45^o. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

Let, AD is the lighthouse and B and C are two ships.

Given, the height of lighthouse AD = 75 m

And the angle of depression of ship C from the top of the lighthouse ∠ CDE = 30^o

Thus, ∠ ACD = 30^o

Again, given, the angle of depression from the ship B to the top of the lighthouse, ∠ BDE = 45^o

Thus, ∠ ABD = 45^o

Thus, Distance, between ships = BC = ?

Now, in Δ ABD

AD = perpendicular ( p )

And AB = base ( b )

And ∠ ABD = 45^o

We know that, tan θ = ^p/_b

⇒ tan 45^o = ^AD/_AB

⇒ 1 = ^{75 m}/_AB

After cross multiplication

⇒ AB = 75 m ------(i)

Now, in Δ ACD

AD = perpendicular ( p )

AC = base ( b )

And angle, ∠ ACD = 30^o

We know that, tan θ = ^p/_b

⇒ tan 30^o = ^AD/_AC

⇒ ¹/_√3 = ^{75 m}/_AC

By cross multiplication

⇒ AC = 75 √3 m

⇒ AB + BC = 75 √3 m

[∵ AC = AB + BC]

By substituting the value of AB from equation (i), we get

75 m + BC = 75 √3 m

⇒ BC = 75 √3 m – 75 m

⇒ BC = 75 (√3 – 1) m

Thus distance between given two ships = 75 (√3 – 1) m Answer

Question: (14) A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60^o. After some time, the angle of elevation reduces to 30^o (see figure). Find the distance traveled by the balloon during the interval.

Solution:

Given, height of the balloon from ground, BM = FN = 88.2 m

Height of the girl AC = BD = 1.2 m

Thus, EN = DM = BM –  AC = 88.2 –  1.2 = 87 m

∠ DCM = 30^o

And, ∠ ECN = 60^o

Thus, distance travelled by balloon, ED = FB = ?

Now, in Δ ECN

We know that, tan θ = ^p/_b

⇒ tan 60^o = ^EN/_CE

⇒ √3 = ^{87 m}/_CE

After cross multiplication

⇒ CE √3 = 87 m

⇒ CE = ^{87 m}/_√3 -----(i)

Now, in Δ CDM

We know that, tan θ = ^p/_b

⇒ tan 30^o = ^DM/_CD

⇒ ¹/_√3 = ^{87 m}/_CD

After cross multiplication

⇒ CD = 87 √3 m

⇒ CE + ED = 87 √3 m

[∵ CD = CE + ED]

After substituting the value of CE from equation (i)

⇒ ⁸⁷/_√3 m + ED = 87 √3 m

⇒ ED = 87 √3 m – ⁸⁷/_√3 m

⇒ ED = ^{87 √3 × √3 – 87}/_√3 m

⇒ ED = ^{87 × 3 –  87}/_√3 m

⇒ ED = ^{87(3 – 1)}/_√3 m

⇒ ED = ^{87 × 2}/_√3 m

⇒ ED = ¹⁷⁴/_√3 m

⇒ ED = ¹⁷⁴/_√3 × ^√3/_√3 m

⇒ ED = ^174√3/₃ m

⇒ ED = 58 √3 m

Thus, distance travelled by balloon is equal to 58 √3 m Answer

Question: (15) A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a cart at an angle of depression of 30^o, which approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60^o. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

Let AB is the tower, and a man sees the car D, the angle of depression of the car at point = 30^o

And when the car reaches point C, the angle of depression becomes 60^o

Given, the time taken to reach the point C from D by car = 6 second

Thus, the time is taken by car to reach the foot of the tower, B from point C = ?

From the figure drawn by the condition given in the question,

∠ ADB = 30^o

And ∠ ACB = 60^o

Now, in Δ ABC

We know that, tan θ = ^p/_b

⇒ tan 60^o = ^AB/_BC

⇒ √3 = ^AB/_BC

After cross multiplication, we get

AB = √3 BC ----(i)

Now, in Δ ABD

We know that, tan θ = ^p/_b

⇒ tan 30^o = ^AB/_BD

⇒ ¹/_√3 = ^AB/_BD

After cross multiplication, we get

BD = √3 AB

⇒ AB = ^BD/_√3 ------(ii)

Now, from equations (i) and (ii)

√3 BC = ^BD/_√3

After cross multiplication, we get

√3 × √3 × BC = BD

⇒ 3BC = BD

⇒ 3BC = BC + CD

[∵ BD = BC + CD]

⇒ CD = 3BC –  BC

⇒ CD = 2BC

⇒ BC = ¹/₂ CD

∵ time taken to cover a distance CD unit by car = 6 second.

∴ time taken to cover a distance of 1 unit by car = ⁶/_CD second

∴ time taken to cover a distance of ¹/₂CD by car = ⁶/_CD × ¹/₂ CD second

= 3 second

Thus, car will reach at the foot of the tower from a given point in 6 seconds. Answer

Question: (16) The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution:

Let DC is the given tower.

And two points A and B angles of elevation are considered.

Given, ∠ DAC and ∠ DBC are complementary.

Let, ∠ DAC = A

Therefore, ∠ DBC = 90^o – A

Again, given, DA = 4 m

And DB = 9 m

Thus, to prove that height of tower = 6m

Now, in Δ DAC

DA = 4 m = base ( b )

And DC = perpendicular ( p )

And, ∠ DAC = A

Now, we know that, tan A = ^p/_b

⇒ tan A = ^CD/_DA

⇒ tan A = ^CD/_{4 m} -----(i)

Now, in Δ DBC

DB = base( b ) = 9 m

And, ∠ DBC = 90^o-A

And CD = perpendicular ( p )

We know that, tan A = ^p/_b

⇒ tan (90^o – A) = ^CD/_DB

⇒ cot A = ^CD/_{9 m} ------(ii)

[∵ tan (90^o – A) = cot A]

After multiplying equations (i) and (ii), we get

tan A × cot A = ^CD/_{4 m} × ^CD/_{9 m}

⇒ tan A × ¹/_{tan A} = ^{CD × CD}/_{4 m × 9 m}

⇒ 1 = ^(CD)2/_{36 m²}

After cross multiplication, we get

(CD)² = 36 m²

⇒ CD = √(36 m²)

⇒ CD = 6 m = height of the tower

Thus, height of tower = 6 m. Proved

MCQs Test

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