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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