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 :









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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



  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. View Hint View Answer Discuss in Forum

    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




  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. View Hint View Answer Discuss in Forum

    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.



  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. View Hint View Answer Discuss in Forum

    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




  1. A vertical pole is 75m high. Find the angle subtended by the pole a point 75 m away from its base.









  1. View Hint View Answer Discuss in Forum

    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°