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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 13 Heights and Distances Ex 13.1

Part – A

**2nd PUC Basic Maths Heights and Distances Ex 13.1 Two Marks Questions and Answers**

Question 1.

The angle of elevation of the top of a tower at a distance 500 metres from its foot is 30°. Find the height of the tower.

Sol.

From the triangle ABC

B be the position of the observer & AC = height of the pole h.

tan 30° = \(\frac{\mathrm{AC}}{\mathrm{AB}}\); \(\frac{1}{\sqrt{3}}=\frac{h}{500}\)

∴ h = \(\frac{500}{\sqrt{3}}=500 \sqrt{3}\) Height of the tower m \(\frac{500 \sqrt{3}}{2} \mathrm{mts}\)

Question 2.

The angle of elevation of the top of a chimney at a distance of 100 metres from a foot is 30°. Find its height.

Answer:

in triangle ABC we have B as the position of observer, AC height of tan 30° = \(\frac{A C}{A B}\) the chimney = h

\(\frac{1}{\sqrt{3}}=\frac{h}{100}\)

\(\mathrm{h}=\frac{100}{\sqrt{3}}=\frac{100 \sqrt{3}}{3} \mathrm{m}\)

Question 3.

From a ship a mast head 40 meters high the angle of depression of a boat is observed to be 45°. Find its distance from the ship.

Answer:

In triangle ABC, we have AC as mast head, AB is distance from the ship

tan 45° = \(\frac{A C}{A B}\)

\(1=\frac{40}{x}\)

⇒ x = 40

∴ Distance from the ship is 40 mts

Question 4.

What is the angle of elevation of the sun when the length of the shadow of a pole is \(\frac{1}{\sqrt{3}}\) times the height of the pole?

Answer:

Let the height of the pole be AC = h & length of the shadow = AB = \(\frac{1}{\sqrt{3}}\) h

From triangle ABC we have tan

tan θ = \(\frac{A C}{A B}\) tan θ = \(\frac{\mathrm{h}}{\frac{1}{\sqrt{3}}}=\sqrt{3}\) = tan 60°

⇒ θ = 30°

Question 5.

Find the angle of elevation of the sun when the shadow of a tower 75 meters high is meters \(25\sqrt{3}\) long.

Answer:

AC = height of the tower = h = 75 mts AB = Shadow of a tower = \(25\sqrt{3}\) from triangle ABC, we have

tan θ = \(\frac{A C}{A B}=\frac{75}{25 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\) = tan 60°

⇒ θ = 60°

Question 6.

A kite flying at a height of h is tied to a thread which is 500m long. Assuming that there is no kink in the thread and it makes an angle of 30° with the ground. Find the height of the kite.

Answer:

Let AC = height of the kite BC = 500 mts From ∆ ABC we have

sin 30° = \(\frac{A C}{B C}=\frac{h}{500}\)

\(\frac{1}{2}=\frac{h}{500} \Rightarrow h=\frac{500}{2}=250 \mathrm{mts}\)

∴ the heights of the kite = 250 mts.

Question 7.

A ladder leaning against a wall makes an angle of 60° with the ground. The foot of the ladder is 6m away from the wall. Find the length of the ladder.

Answer:

Let AC be the ladder from triangle ABC we have

cos 60° = \(\frac{A B}{B C}=\frac{6}{B C}\)

\(\frac{1}{2}=\frac{6}{\mathrm{BC}} \Rightarrow \mathrm{BC}=12 \mathrm{mts}\)

∴ The length of the ladder = 12 mts.

Question 8.

Find the angle of elevation of the sun’s rays from a point on the ground at a distance of \(3 \sqrt{3}m\) , from the foot of tower 3m high.

Answer:

Let AC be the height of the pole AC = 3mts & AB = \(3 \sqrt{3}m\) mts From the triangle ABC we have

tan θ = \(\frac{A C}{A B}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}}\) = tan 30 ⇒ = 30°

∴ The angle of elevation is 30°

Part – B

**2nd PUC Basic Maths Heights and Distances Ex 13.1 Four or Five marks questions and answers**

Question 1.

The angles of elevation of the top of a tower from the base and the top of a building are 60° and 45° The building is 20 meters high. Find the height of the tower.

Answer:

Let AD be the tower & CE be the building. Let the distance between the tower and building be x in ∆ ABC,

tan 45° = \(\frac{A B}{B C}=\frac{h}{x}\)

\(1=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{x}=\mathrm{h}-(1)\)

Question 2.

The shadow of a tower standing on a level plane is found to be 50 meters longer when sun’s altitude is 30°. Than when it is 60°. Find the height of the tower.

Answer:

Let AB be the tower In triangle ABC, tan 60° = \(\frac { h }{ x }\)

\(\sqrt{3}=\frac{h}{x}\)

\(h=\sqrt{3 x}\)

In triangle ABD, tan 30° = \(\frac{h}{50+x}\)

50 + x = \(\sqrt{3}\)

50 + x = \(\sqrt{3} \cdot \sqrt{3} x\) ∵ h = \(\sqrt{3} x\)

50 = 3x – x ⇒ 2x = 50 ⇒ x = 25 mts

∴ Height of the tower h = 25\(\sqrt{3}\) mts

Question 3.

An Aeroplane when flying at a heights of 2000 meters passes vertically above another plane at an instant, when their angles of elevation from the same point of observation are 60° and 45° respectively. Find the distance between the Aeroplanes.

Answer:

Let A & B are Aeroplanes Let h be the distance between the aeroplanes

From triangle ACD

From triangle BCD

Question 4.

From a point on the line joining the feet of two poles of equal heights, the angles of elevation of the tops of the poles are observed to be 30° and 60°. If the distance between the poles is a Find (i) the height of the poles (ii) the position of the point of observation.

Answer:

Let AB & DE are two poles

∴ Height of the poles is \(\frac{\sqrt{3}}{4} a\)

Position of c from B is x = \(\sqrt{3 h}\)

∴ position of the point of observation = \(\frac{\sqrt{3} \cdot \sqrt{3} a}{4}=\frac{3 a}{4}\)

Question 5.

The angles of elevation of the top of a tower from two points distant a and b (a < b) from its foot and the same straigth line from it are 30° and 60°. Show that the height of the tower is \(\sqrt{a b}\).

Answer:

Let CD is the tower from rt angled triangle ACD

we have tan 30 = \(\frac { h }{ a }\) ⇒ h = \(\frac{a}{\sqrt{3}}\) …(1)

From rt angled triangle BCD

tan 60° = \(\frac { h }{ b }\)

\(\sqrt{3}=\frac{h}{b} \Rightarrow h=\sqrt{3} b-(2)\)

From 1 & 2 we have

Question 6.

A flag staff stands upon the top of a building. At a distance of 20 meters the angles of elevation of the top of the flag staff and building are 60° and 60° respectively. Find the height of the flag staff.

Answer:

Let BC = Building

AB = flagstaff

From triangle ACD we have

tan 60° = \(\frac{A C}{C D}=\frac{A B+B C}{20}=\sqrt{3}\)

AB + BC = 20 \(\sqrt{3}\) …. (1)

In triangle BCD, tan 30° = \(\frac{B C}{C D} ; \quad \frac{1}{\sqrt{3}}=\frac{B C}{20}\)

BC = \(\frac{20}{\sqrt{3}}\) …. (2)

Question 7.

From the top of a cliff, the angles of depression of two boats in the same vertical plane as the observer are 30° and 45°. If the distance between the boats is 100 meters, find the height of the cliff.

Answer:

Let the height of the cliff be h A & B are two boats

From triangle DCA

tan 45° = \(\frac{D C}{A C}=\frac{h}{x}\)

\(1=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{h}=\mathrm{x}\) ….. (1)

From triangle BCD

Question 8.

From a point A due north of the tower, the elevation of the top of the tower is 60°. From a point B due south, the elevantion is 45°, if AB = 100 metres. Show that the height of the tower is 50 \(\sqrt{3}(\sqrt{3}-1)\) meters.

Answer:

A and B are positions of observation AB = 100 mts and CD is the height of tower and CD = h

From rt angled triangle ACD

tan 60° = \(\frac{\mathrm{CD}}{\mathrm{AD}}\)

\(\sqrt{3}\) = \(\frac{h}{A D} \Rightarrow A D=\frac{h}{\sqrt{3}}\) …..(1)

From rt angled triangle BCD

tan 45° = \(\frac{\mathrm{CD}}{\mathrm{BD}}\)

\(1=\frac{h}{B D} \Rightarrow B D=h\) …. (2)

Question 9.

A person at the top of a hill observes that the angles of depression of two consecutive kilometers stones on a road leading to the foot of the hill and in the same vertical plane containing the position of the observer are 30° and 60°. Find the height of the hill.

Answer:

Let CD be the hill and CD = h mts

In ∆ ADC, we have tan 60° = \(\frac{C D}{A D} \Rightarrow \sqrt{3}=\frac{h}{x}\)

\(\Rightarrow \quad x=\frac{h}{\sqrt{3}} \Rightarrow h=\sqrt{3} x\) …. (1)

From 1 and 2 we get

1 + x = \(\sqrt{3} \cdot \sqrt{3} x\) (h = \(\sqrt{3} x\))

1 + x = 3x

1 = 2x ⇒ x = 5 \(\frac{1}{2}\)

The height of the hill = h = \(\sqrt{3} x=\sqrt{3} \cdot \frac{1}{2}=\frac{\sqrt{3}}{2} \mathrm{mts}\) mts

Question 10.

The angles of elevation of the top of a tower from the base the top of a building are 60° & 45° the building is 32 meters high find the height of the tower

Answer:

Let CD is the tower

AB is the house = 32 meters

From the rt angled triangle ACE

tan 45° = \(\frac{C E}{A E}=\frac{C E}{B D}=1\)

⇒ CE = BD …. (1)

Again from rt angled triangle ABD

tan 30° = \(\frac{A B}{B D}=\frac{32}{B D}=\frac{1}{\sqrt{3}}\)

⇒ BD = \(32 \sqrt{3}\)

From 1 and 2 we get CE = \(32 \sqrt{3}\) mts

∴ Height of the tower = h = DE + CE = 32 + \(32 \sqrt{3}\) = -32(1 + \(\sqrt{3}\))mts

Question 11.

The angle of elevation of a tower from a point on the ground is 30°. At a point on the horizontal line passing through the foot of the tower and 100 metres nearer it, the angle of elevation is found to be 60°. Find the height of the pole.

Answer:

Let CD = Tower

From a rt angled triangle ACD

tan 30° = \(\frac{\mathrm{CD}}{\mathrm{AC}}\)

\frac{1}{\sqrt{3}}=\(\frac{h}{100+x}\)

100+ x = \(\mathrm{h} \sqrt{3}\) …. (1)

From triangle BCD

tan 60° = \(\frac{\mathrm{CD}}{\mathrm{CB}}\)

\(\sqrt{3}=\frac{h}{x} \Rightarrow h=\sqrt{3} x\) …. (2)

From 1 & 2 we get

100 + x = \(\sqrt{3} \cdot \sqrt{3} x\)

100 = 3x – x ⇒ 2x = 100 ⇒ x = 50 :

∴ Height of the tower = h = \(\sqrt{3 x}\) = \(50\sqrt{3}\) mts

Hence Distance of the first point from the tower ₹ 100 + x = 100 + 50 = 150 mts

Question 12.

A Person is at the top of a tower 75 feet high from there he observes a vertical pole and finds the angles :: of depressions of the top and the bottom of the pole which are 30° and 60° respectively. Find the height of the pole.

Answer:

Let AB = Tower, CD = Pole = h

From triangle ABC we get

tan 60° = \(\frac{A B}{A C} \Rightarrow \frac{75}{A C}=\sqrt{3}\)

⇒ AC = \(\frac{75}{\sqrt{3}} ….. (1)\) …. (1)

From triangle BDE

⇒ AC = \(\sqrt{3}(75-h)\)

From 1 and 2 we get

\(\frac{75}{\sqrt{3}}=\sqrt{3}(75-h) \Rightarrow 75=(\sqrt{3})^{2}(75-h)\)

75 = 225 – 3h 3h ⇒ 150 ⇒ h = 50

∴ Height of the pole -50ft