Heights and Distance – Case Based Questions

Q 1 :

One evening, Kaushik was in a park where children were playing cricket. Birds were singing in a nearby tree that was 80 m tall. He observed a bird on the tree at an angle of elevation of $45 °$ .

When a sixer was hit, the ball flew through the tree, scaring the bird away. 2 seconds later, Kaushik noticed that the bird was flying at the same height but at an angle of elevation of $30 °$ . Meanwhile, the ball was also flying towards him at the same height with an angle of elevation of $60 °$ .

Based on the above information, answer the following questions:

(i) At what distance from the foot of the tree was he observing the bird sitting on the tree?

80 m
40 m
55 m
50 m

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(1)

$I n ∆ A B C t a n 45 ° = \frac{A B}{C B} 1 = \frac{80}{C B} \Rightarrow C B = 80 m$

Q 2 :

One evening, Kaushik was in a park where children were playing cricket. Birds were singing in a nearby tree that was 80 m tall. He observed a bird on the tree at an angle of elevation of $45 degree$ .

When a sixer was hit, the ball flew through the tree, scaring the bird away. 2 seconds later, Kaushik noticed that the bird was flying at the same height but at an angle of elevation of $30 degree$ . Meanwhile, the ball was also flying towards him at the same height with an angle of elevation of $60 degree$ .

Based on the above information, answer the following questions:

(ii) (a) How far did the bird fly in the mentioned time?

$80 (\sqrt{3} - 2) m$
$80 (\sqrt{3} - 1) m$
$80 (\sqrt{2} - 1) m$
$80 (\sqrt{3} + 2) m$

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(2)

$I n ∆ C E D, t a n 30 ° = \frac{D E}{C E} \frac{1}{\sqrt{3}} = \frac{80}{C E} \Rightarrow C E = 80 \sqrt{3} m$

Distance the bird flew:

$A D = B E = C E - C B = 80 \sqrt{3} - 80 = 80 (\sqrt{3} - 1) m$

Q 3 :

One evening, Kaushik was in a park where children were playing cricket. Birds were singing in a nearby tree that was 80 m tall. He observed a bird on the tree at an angle of elevation of $45 degree$ .

When a sixer was hit, the ball flew through the tree, scaring the bird away. 2 seconds later, Kaushik noticed that the bird was flying at the same height but at an angle of elevation of $30 degree$ . Meanwhile, the ball was also flying towards him at the same height with an angle of elevation of $60 degree$ .

Based on the above information, answer the following questions:

(ii) (b) After hitting the tree, how far did the ball travel horizontally in the sky when Kaushik saw the ball?

$80 (1 - \frac{1}{\sqrt{3}}) m$
$80 (1 - \frac{2}{\sqrt{3}}) m$
$80 (1 + \frac{1}{\sqrt{3}}) m$
$80 (1 + \frac{- 1}{\sqrt{3}}) m$

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(1)

$I n ∆ C G F t a n 60 ° = \frac{F G}{C G} \sqrt{3} = \frac{80}{C G} \Rightarrow C G = \frac{80}{\sqrt{3}} D i s \tan c e t h e b a l l t r a v e l l e d a f t e r h i t t i n g t h e t r e e : F A = G B = C B - C G = 80 - \frac{80}{\sqrt{3}} = 80 (1 - \frac{1}{\sqrt{3}}) m$

Q 4 :

One evening, Kaushik was in a park where children were playing cricket. Birds were singing in a nearby tree that was 80 m tall. He observed a bird on the tree at an angle of elevation of $45 degree$ .

When a sixer was hit, the ball flew through the tree, scaring the bird away. 2 seconds later, Kaushik noticed that the bird was flying at the same height but at an angle of elevation of $30 degree$ . Meanwhile, the ball was also flying towards him at the same height with an angle of elevation of $60 degree$ .

Based on the above information, answer the following questions:

(iii) What is the speed of the bird in m/min if it had flown $20 (\sqrt{3} + 1) m ?$

$600 (\sqrt{3} - 1) m / \min$
$600 (\sqrt{2} + 1) m / \min$
$500 (\sqrt{3} + 1) m / \min$
$600 (\sqrt{3} + 1) m / \min$

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(4)

Speed of the bird:

$Speed = \frac{Distance}{Time} = \frac{20 (\sqrt{3} + 1)}{2} m / \sec Convert to m / \min by multiplying by 60 : = \frac{20 (\sqrt{3} + 1)}{2} \times 60 = 600 (\sqrt{3} + 1) m / \min$

Q 5 :

Radio towers are essential for transmitting various communication services, including radio and television. These towers either function as antennas themselves or support one or more antennas on their structure. Following this concept, a radio station tower was constructed in two sections, A and B. The tower is supported by wires extending from a point O.

Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of section B is $30 degree$ and the angle of elevation of the top of section A is $45 degree.$

$U s e \sqrt{3} = 1.73$

Based on the above information, answer the following questions:

(i) Find the length of the wire from point O to the top of Section B

$41.52 cm$
$- 41.52 cm$
$41.50 cm$
$40.52 cm$

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(1)

Let OB be the length of the wire from O to top of section B

$I n ∆ O P B c o s 30 ° = \frac{O P}{O B} \frac{\sqrt{3}}{2} = \frac{36}{O B} O B = \frac{72}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{72 \sqrt{3}}{3} = 24 \sqrt{3} cm OB = 24 \times 1.73 = 41.52 cm$

Q 6 :

Radio towers are essential for transmitting various communication services, including radio and television. These towers either function as antennas themselves or support one or more antennas on their structure. Following this concept, a radio station tower was constructed in two sections, A and B. The tower is supported by wires extending from a point O.

Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of section B is $30 degree$ and the angle of elevation of the top of section A is $45 degree.$

$U s e square root of 3 equals 1.73$

Based on the above information, answer the following questions:

(ii) (a) Find the distance AB.

$15.20 cm$
$15.24 cm$
$14.24 cm$
$13.24 cm$

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(2)

$I n ∆ A P O : t a n 45 ° = \frac{A P}{O P} 1 = \frac{A P}{36} \Rightarrow A P = 36 cm In ∆ BPO \tan 30 ° = \frac{PB}{OP} \frac{1}{\sqrt{3}} = \frac{PB}{36} PB = \frac{36}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{36 \sqrt{3}}{3} = 12 \sqrt{3} cm PB = 12 \times 1.73 = 20.76 cm Now, distance : AB = AP - PB = 36 - 20.76 = 15.24 cm$

Q 7 :

Radio towers are essential for transmitting various communication services, including radio and television. These towers either function as antennas themselves or support one or more antennas on their structure. Following this concept, a radio station tower was constructed in two sections, A and B. The tower is supported by wires extending from a point O.

Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of section B is $30 degree$ and the angle of elevation of the top of section A is $45 degree.$

$U s e square root of 3 equals 1.73$

Based on the above information, answer the following questions:

(ii) (b) Find the area of $△ O P B$ .

$375.68 {cm}^{2}$
$363.68 {cm}^{2}$
$373.68 {cm}^{2}$
$- 373.68 {cm}^{2}$

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(3)

$A r e a o f ∆ O P B : = \frac{1}{2} \times O P \times P B = \frac{1}{2} \times 36 \times 12 \sqrt{3} = 18 \times 12 \sqrt{3} = 18 \times 12 \times 1.73 = 373.68 {cm}^{2}$

Q 8 :

Radio towers are essential for transmitting various communication services, including radio and television. These towers either function as antennas themselves or support one or more antennas on their structure. Following this concept, a radio station tower was constructed in two sections, A and B. The tower is supported by wires extending from a point O.

Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of section B is $30 degree$ and the angle of elevation of the top of section A is $45 degree.$

$U s e square root of 3 equals 1.73$

Based on the above information, answer the following questions:

(iii) Find the height of Section A from the base of the tower.

36 cm
34 cm
30 cm
25 cm

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(1)

$S i n c e A P = 36 c m (f r o m p a r t (i i)),$

Hence, the height of section A from the base of the tower is 36 cm.

Q 9 :

Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS, a radio navigation system, helps to locate our position on earth with the help of satellites. A guard, stationed at the top of a 240 m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes (tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°.

(Lighthouse of Mumbai Harbour — Picture credits: Times of India Travel)

(i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower

$240 \sqrt{3} m$
$240 \sqrt{2} m$
$- 240 \sqrt{3} m$
$235 \sqrt{3} m$

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(1)

Let TR be the tower, observer at point T, P be the boat and PR = x.

From the figure,

$∠ M T P = ∠ T P R = 30 ° t a n 30 ° = \frac{T R}{P R}$

$\frac{1}{\sqrt{3}} = \frac{240}{x} \Rightarrow x = 240 \sqrt{3} m$

$H e n c e, t h e b o a t i s 240 \sqrt{3} m$

Q 10 :

Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS, a radio navigation system, helps to locate our position on earth with the help of satellites. A guard, stationed at the top of a 240 m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes (tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°.

(Lighthouse of Mumbai Harbour — Picture credits: Times of India Travel)

(ii) After 10 minutes, the guard observed that the boat was approaching the tower and its distance from the tower is reduced by $240 (\sqrt{3} - 1) m . .$ He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower?

$35 °$
$45 °$
$40 °$
$25 °$

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(2)

Let Q be the position of the boat after 10 minutes.

$P Q = 240 (\sqrt{3} - 1) m Now, QR = PR - PQ$

$= 240 \sqrt{3} - 240 (\sqrt{3} - 1) = 240 m$

$Let θ be the angle of depression when the boat is at point Q .$

$∠ MTQ = ∠ TQR = θ$

$In ∆ TRQ : tanθ = \frac{TR}{QR} = \frac{240}{240} = 1$

$\Rightarrow tanθ = \tan 45 ° \Rightarrow θ = 45 °$

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