Height and Distance


  1. The angles of depression of two ships from the top of a light house are 45° and 30° towards east. if the ships are 200 m apart, find the height of the light house .









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    Let us draw a figure below as per given question.
    Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ∠ADC = 30° ∠ACB = 45° and BC = x meter (say)


    Correct Option: D

    Let us draw a figure below as per given question.
    Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ∠ADC = 30° ∠ACB = 45° and BC = x meter (say)
    Now from right triangle ABC,
    tan 45° = h/x ⇒ 1 = h/x
    ∴ x = h
    Again from right triangle ABD,
    tan 30° = h/( 200 + x )
    ⇒ 1/√3 = h/( 200 + x )
    Since x = h , we will get.
    ⇒ 1/√3 = h/(200 + h )
    ⇒ (200 + h ) = h X √3
    ⇒ h X √3 - h = 200
    ⇒ h x 1.732 - h = 200
    ⇒ h(1.732 - 1) = 200
    ∴ h = 200/0.732 = 273.2 m = 273 m



  1. At a point on a level plane a tower subtends an angle Θ and a flag-staff a ft. in length at the top of the tower subtends an angle Θ . The height of the tower is :









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    Let us draw a figure below as per given question.
    Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that OAP = Θ and PAQ = Φ
    In ΔOAP and ΔOAQ, Apply the formula
    TanΘ = P/B = Perpendicular distance / Base distance
    Proceed further and solve the given question by applying the above formula.


    Correct Option: B

    Let us draw a figure as per given question.
    Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that OAP = Θ and PAQ = Φ
    In ΔOAP use the formula
    TanΘ = P/B = Perpendicular distance / Base distance
    ⇒ TanΘ = OP/OA
    ⇒ OA = OP CotΘ
    Put the value of OP in above equation, we will get
    ⇒ OA = hCotΘ ................ (1)
    Now from triangle ΔOAQ,
    ⇒ Tan(Θ + Φ) = OQ/OA
    ⇒ OA = OQ Cot(Θ + Φ)
    ⇒ OA = (h + a) Cot(Θ + Φ) .............(2)
    from equation (1) and (2), We will get
    hCotΘ = (h + a) Cot(Θ + Φ)
    ⇒ CotΘ / Cot(Θ + Φ) = (h + a) / h
    ⇒ CotΘ / Cot(Θ + Φ) = h / h + a / h
    ⇒ CotΘ / Cot(Θ + Φ) = 1 + a / h
    ⇒ (CotΘ / Cot(Θ + Φ)) - 1 = a / h
    ⇒ (CotΘ - Cot(Θ + Φ) ) / Cot(Θ + Φ) = a / h
    h = aCot (Θ + Φ) / CotΘ - Cot(Θ + Φ)
    After converting Cot into Sin and Cos, We will get
    h = aSinΘCos(Θ + Φ)/SinΦ




  1. A 6 ft-tall man finds that the angle of elevation of the top of a 24 ft-high pillar and the angle of depression its base are complementary angles. Then the distance between pillar and man is ?









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    Let us draw a figure below from the given question.
    Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively.
    Here ∠ ACE = ∝ ∠ CBD = 90° - ∝


    Correct Option: C

    Let us draw a figure below from the given question.
    Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively.
    Here ∠ ACE = ∝ ∠ CBD = 90° - ∝
    Now from right triangle CDB,
    BD = 6 cot (90° - ∝) = 6 tan ∝
    ∴ tan ∝ = BD/6 ...............(i)
    From right triangle ACE,
    tan ∝ =18/EC = 18/BD ...............(ii) (∵ EC = BD)
    Now from (i) and (ii) we get,
    BD/6 = 18/BD
    ⇒ BD2 = 18 x 6
    ∴ BD = 6√3 ft.
    Hence, distance between pillar and man = 6√3 ft



  1. Two ships leave a port at the same instant. One sails at 30km/hr in the direction N 32° E while the other sails at 20 km/hr in the direction S 58° E. After two hours the ships are distant from each other by.









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    Let us draw a figure below as per given question.
    Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.


    Correct Option: C

    Let us draw a figure below as per given question.
    Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.
    Now, ∠NOA = 32° and ∠SOB = 58° ,
    Then, ∠AOB = 180° - (32° + 58°) = 90°
    Since ΔAOB is a right triangle in which OA = 2 x 30 = 60 km and OB = 2 x 20 = 40 km
    Since, AB = √OA2 + OB2
    = √(60)2 + (40)2
    = √5200
    = 20√13




  1. ABC is a triangular park with AB = AC = 100 meters. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower of A and B are cot-1 (3.2) and cosec-1(2.6) respectively. The height of the tower in meters is ?









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    Let us draw a figure as per given question.
    Given ∝ = cot-1 (3.2) and β = cosec-1(2.6)
    In triangle ΔPAD, AD = h cot ∝
    In triangle ΔPBD, BD = h cot β


    Correct Option: B

    Let us draw a figure below as per given question.
    Given ∝ = cot-1 (3.2) and β = cosec-1(2.6)
    In triangle ΔPAD, AD = h cot ∝
    In triangle ΔPBD, BD = h cot β
    In triangle Δ ABD,
    AB2 = AD2 + BD2
    = h2(cot2∝ + cot2 β )
    ⇒ 1002 = h2{ cot2 ∝ + (cosec2 β - 1) }
    ⇒ 1002= h2 { (3.2)2 + (2.6)2 - 1} = 16h2
    ⇒ h = 25 m.