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Height and Distance

  1. 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 :
    1. 14.8 m
    2. 24.8 m
    3. 6.2 m
    4. 12.4 m
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    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


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  1. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°, when he retires 40 meters from the bank then he find the angle to be 30°. Then the breadth of the river is ?
    1. 40 m
    2. 60 m
    3. 20 m
    4. 30 m
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    Let us draw a figure below as per given question.
    Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and OAP = 60°. When the person retires to the position B, then AB = 40 meter and OBP = 30°
    Use the trigonometry formula to solve the given question.


    Correct Option: C

    Let us draw a figure below as per given question.
    Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and OAP = 60°. When the person retires to the position B, then AB = 40 meter and OBP = 30°
    Let us assume OA(Breadth of the river) = x meter and height of tree OP = h meter
    In ΔOAP, Use the trigonometry formula
    Tan60° = P/B = Perpendicular distance / Base distance
    Tan60° = OP / OA
    ⇒ OP = OA Tan60°
    Put the value of OP and OA, We will get
    h = x√3 ..............(1)
    Now in the triangle ΔOBP
    Tan30° = OP / OB
    ⇒ OP = OB Tan30°
    ⇒ OP = (x + 40)/√3
    h = (x + 40)/√3 ...................(2)
    From Equation (1) and (2), We will get
    ⇒ (x + 40)/√3 = x√3
    ⇒ (x + 40) = x√3 X √3
    ⇒ (x + 40) = 3x
    ⇒ 3x - x = 40
    ∴ x = 20 m


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  1. A man is observing from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 m from the tower. After 5 second, the angle of depression becomes 30°, Find the speed of the boat, assuming that it is running in still water.
    1. 30 km/hr.
    2. 31.5 km/hr
    3. 33 km/hr
    4. 34 km/hr
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    Let us draw the figure from the given question.
    Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ∠ACB = 45° and ∠ADB = 30°


    Correct Option: B

    Let us draw the figure from the given question.
    Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ∠ACB = 45° and ∠ADB = 30°
    Now from right triangle ABC,
    tan 45° = h/60
    ⇒ 1 = h/60
    ∴ h = 60 m;
    Again from right triangle ABD;
    tan 30° = h/(x + 60)
    ⇒ 1/√3 = 60/(x + 60)
    ⇒ x + 60 = 60√3
    ∴ x = 60(1.73 - 1) = 43.8 meter
    Hence, speed of boat = 43.8/5 m/s = 43.8/5 x 18/5 = 31.5 km/hr.


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  1. From a point P on a level ground, the angle of elevation of the top of a tower is 30°, if the tower is 100 m high, find the distance of point P from the foot of the tower.
    1. 100 m
    2. 173 m
    3. 200 m
    4. 273 m
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    Let us draw a figure from given question.
    Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°


    Correct Option: B

    Let us draw a figure from given question.
    Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°
    From right triangle ABP,
    BP = 100 cot 30°
    ⇒ BP = 100 √3
    ⇒ BP = 100 X 1.73
    ⇒ BP = 173 meter


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  1. A vertical pole is 75m high. Find the angle subtended by the pole a point 75 m away from its base.
    1. 30°
    2. 45°
    3. 60°
    4. 90°
  1. View Hint | View Answer | Workspace | Discuss In Forum | Report

    Let us draw a figure below as per given question.
    Let AB = 75 m be the height of pole and C is a point on the ground such that BC = 75m


    Correct Option: B

    Let us draw a figure below as per given question.
    Let AB = 75 m be the height of pole and C is a point on the ground such that BC = 75m
    Now, from right triangle ABC.
    tan ∝ = AB/BC = 75/75
    ⇒ tan∝ = 1
    ∴ ∝ = 45°


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