Height and Distance
 The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. The height of the tower is :

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Let us draw the figure from the given question.
Let us assume the angle of elevation ∠MŌP = 45°.
Given : MO = 20 m
In triangle MOP ,tan45° = PM MO Clearly, h = tan45 = 1 20
Correct Option: B
From the given figure , we can see that
Let us assume the angle of elevation ∠MŌP = 45°. and The height of the tower = h
Given : MO = 20 m
In triangle MOP ,tan45° = PM MO ⇒ h = tan45° = 1 20
∴ h = 20 m
 From the top of a 25 m high, cliff the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find out the height of the tower.

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Let us draw the figure from the given question.
Let, AB be the cliff and CD be the tower. From B, draw BE ⊥ CD.
Let us assume the angle of elevation ∠ACB = ∠EBD = α.
Correct Option: C
From the figure , we can see that
Let, AB be the cliff and CD be the tower. From B, draw BE ⊥ CD.
Given : AB = 25 m
In triangle EBD and ACB , we haveDE = tan α and AB = tan α BE AC ⇒ DE = AB BE AC
∴ DE = AB ( ∵ BE = AC)
∴ CD = CE + DE = AB + AB = 2AB = 2 x 25 = 50 m.
 The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is :

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Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Correct Option: B
Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Cos 60° = PQ / PR
⇒ 1 / 2 = 12.4 / PR
⇒ PR = 2 × 12.4 = 24.8 m
 A tower stands at the end of a straight road. The angles of elevation of the top of the tower from two points on the road 500 m apart are 45° and 60°, respectively. Find out the height of the tower.

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Let us draw the figure from the given question.
Given , CD = 500 m , θ_{1} = 45° and θ_{2} = 60°
We know that(cot 45°  cot 60°) = CD AB
⇒ CD = AB (cot 45°  cot 60°)⇒ AB = CD (cot 45°  cot 60°)
Correct Option: A
From given figure , we can see that
Given , CD = 500 m , θ_{1} = 45° and θ_{2} = 60°
We know that(cot 45°  cot 60°) = CD AB
⇒ CD = AB (cot 45°  cot 60°)⇒ AB = CD (cot 45°  cot 60°) ⇒ AB = 500 = 500 √3 m. 1  1 √3  1 √3
 When the sun is 30° above the horizontal, the length of shadow cast by a building 50 m high is :

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Let us draw the figure from the given question.
Let, AB be the building and AC be its shadow.
Then, height of building AB = 50 m and θ = 30°.
Correct Option: B
Let us draw the figure from the given question.
Let, AB be the building and AC be its shadow.
Then, height of building AB = 50 m and θ = 30°.In triangle ACB , cotθ = AC AB ∴ AC = cot 30° = √3 ⇒ AC = √3 AB 50
⇒ AC = 50 √3 m.
Hence , the length of shadow cast by a building 50 m high is 50 √3 m .