class 10 maths Some Applications of Trigonometry questions with solutions

Q 1 :
A person/observer on the sea coast observes two ships in the sea, both the ships are in same straight path one behind the other.

If the observer is on his building of height 20 meters (including observer) and he observes the angle of depression of two ships as 45° and 60° respectively.

On the basis of above information answer the following questions.

(i) If a person observes a ship whose angle of depression is 60° then how much distance is the ship away from the building?

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(11.55)

$\tan 60 ° = \frac{O C}{A C} \Rightarrow \sqrt{3} = \frac{20}{A C} \Rightarrow A C = \frac{20}{\sqrt{3}} = \frac{20 \sqrt{3}}{3} m = 11.55 m$

Q 2 :
A person/observer on the sea coast observes two ships in the sea, both the ships are in same straight path one behind the other.

If the observer is on his building of height 20 meters (including observer) and he observes the angle of depression of two ships as 45° and 60° respectively.

On the basis of above information answer the following questions.

(ii) If a person observes another ship whose angle of depression is 45° then how much distance that ship is away from the building?

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(20)

$\tan 45 ° = \frac{O C}{B C} \Rightarrow 1 = \frac{20}{B C} \Rightarrow B C = 20 m$

Q 3 :
A person/observer on the sea coast observes two ships in the sea, both the ships are in same straight path one behind the other.

If the observer is on his building of height 20 meters (including observer) and he observes the angle of depression of two ships as 45° and 60° respectively.

On the basis of above information answer the following questions.

(iii) At a time when a person observes two ships whose angle of depressions are 60° and 45° the distance between the ships is (in meter).

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(8.45)

Distance between two ships 20 m = BC – AC = 20 – 11.55 = 8.45 m

Q 4 :
A person/observer on the sea coast observes two ships in the sea, both the ships are in same straight path one behind the other.

If the observer is on his building of height 20 meters (including observer) and he observes the angle of depression of two ships as 45° and 60° respectively.

On the basis of above information answer the following questions.

(iv) If a person observes the ship whose angle of depression changes from 60° to 30° then how far be ship from the building if the observer is at 20 m of height (including him)?

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(34.6)

$\tan 30 ° = \frac{O B}{O A} \Rightarrow \frac{1}{\sqrt{3}} = \frac{20}{O A} \Rightarrow O A = 20 \sqrt{3} m = 34.6 m$

Q 5 :
Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely X and Y have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point O is fixed in front of the fire engine.

The ladder inclined at an angle 60° to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of 45° with the ground. Both the buildings along with the foot of the ladder, fixed at ‘O’ are in a straight line.

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Based on the above given information, answer the question :

Q. Find the length of the ladder.

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(24)

In $∆ O A P,$ $\cos e c 60^{0} = \frac{O P}{12 \sqrt{3}} = \frac{2}{\sqrt{3}}$

$\Rightarrow$ $O P = 24 m$

$∴$ Length of ladder is 24 m

Q 6 :
Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely X and Y have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point O is fixed in front of the fire engine.

The ladder inclined at an angle 60° to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of 45° with the ground. Both the buildings along with the foot of the ladder, fixed at ‘O’ are in a straight line.

[IMAGE]

Based on the above given information, answer the question :

Q. Find the distance of the building Y from point ‘O’, i.e. OA.

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(12)

In $∆ O A P, \cot 60^{0} = \frac{O A}{12 \sqrt{3}} = \frac{1}{\sqrt{3}}$

$\Rightarrow O A = 12 m$

$∴$ the distance of the building Y from point O ie.,OA is 12 m

Q 7 :
Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely X and Y have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point O is fixed in front of the fire engine.

The ladder inclined at an angle 60° to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of 45° with the ground. Both the buildings along with the foot of the ladder, fixed at ‘O’ are in a straight line.

[IMAGE]

Based on the above given information, answer the question :

Q. Find the horizontal distance between the two buildings.

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( $12 (1 + \sqrt{2})$ )

OP = OR = 24 m

$∴$ In $∆ O C R, \cos 45^{0} = \frac{O C}{24} = \frac{1}{\sqrt{2}}$

$\Rightarrow O C = 12 \sqrt{2} m$

$∴$ distance between two buildings = OA + OC

$= (12 + 12 \sqrt{2})$ m or $12 (1 + \sqrt{2})$ m

Q 8 :
Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely X and Y have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point O is fixed in front of the fire engine.

The ladder inclined at an angle 60° to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of 45° with the ground. Both the buildings along with the foot of the ladder, fixed at ‘O’ are in a straight line.

[iMAGE]

Based on the above given information, answer the question :

Q. Find the height of the building X.

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( $12 \sqrt{2}$ )

OP = OR = 24 m

$∴$ In $∆ O C R, \sin 45^{0} = \frac{R C}{24} = \frac{1}{\sqrt{2}}$

$\Rightarrow R C = 12 \sqrt{2} m$

$∴$ height of building X is $12 \sqrt{2}$ m

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