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    NCERT Solutions for Class 10 Maths – Some Applications of Trigonometry

    1. If sun’s elevation is 45°. Find the length of the shadow of a pole of height 8 m.

    In rt. Δ PQR, B=90° and QPR=45°

    1

    ∴ tan 45° = QRPQ

    1 = 8PQ

    AB=81 = 8m

     

    2. A pole 8 metre high cast a shadow 8 m on the ground. Find the sun’s elevation.

    In rt. Δ PQR, B=90°

    2

    ∴ PQ = 8 m

    QR = 8 m

    Let QPR=θ

    ∴ tan θ = QRPQ

    = QRPQ = 1

    θ = 45°

     

    3. If the angle of elevation of the top of a hill from two points at distances p and q from the base and in the same straight line with it are complementary. Find the height of the hill.

    Let   PQ = h be the height of the hill,

    QR = b, QS = a, such that PSQ=θ

    PRQ=90°θ

    3

    In rt. Δ PQS, we have

    ha=tan θ

    h = a tan θ          …..(1)

    In rt. Δ PQR, we have

    hb=tan(90°θ) = cot θ

    h = b cot θ         …..(2)

    Multiplying (1) and (2), we get

    h2 = ab tan θ cot θ = ab

    h=ab−−√

     

    4. From the figure, find the angle of depression of point Z from the point R.

    4

    YRS=XYR (alternative.int. a )

    YRS = 60°

    ZRS+YRZ=60° ZRS+30°=60° ZRS=30°

     

    5. Find the angle of elevation of the moon when the shadow of a tree h metres high is 3–√h metres long.

    5

    The height of the pole be h m and length of its shadow is 3–√h

    Let θ be the elevation of the sun.

    Consider rt.agl Δ ABC

    tan θ = h3h=13

    = tan 30°

       θ = 30°

     

     6. A ladder 14 metres long just reaches the top of a wall. If the ladder makes an angle of 45° with the wall, find the height of the wall.

    Here, length of the ladder is 14m and angle of elevation is 90°-45°=45°.

    Let h m be the height of the wall

    6

    Consider rt.agl Δ PQR

    QRPR = sin 30°

    h14=22

    h = 1422 m

    ∴ The length of the wall is 72–√

     

     7. An observer 1.8 m tall is 20.8 m away from a tower 22.6 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

    7

    Here, PQ be the observer and RS be the tower of height 24 m.

    PQ = TS = 1.8 m

    RT = RS – TS

    = 22.6 – 1.8

    = 20.8 m

    Now, Consider rt.agl ΔPTR, we have

    RTAT = tan θ

    20.820.8 = tan θ

    1 = tan θ

    tan 45° = tan θ

    θ =45°

     

    8.From a balloon vertically above a straight road, the angles of depression of two cars at an instant is found to be 45° and 60°. If the car are 100 m apart, find the height of the balloon.

    Let the height of the balloon at P be h metres. Let A and B be the two cars. Thus, AB = 100m.

    8

    ∠ PAQ= 45° and ∠ PBQ=60°

    Now, Consider rt.agl ΔAQP,

    PQAQ = tan 45°

             PQAQ = 1

             PQ = AQ= h m

    Now, Consider rt.agl ΔPBQ,

    PQBQ = tan 60°

    PQBQ = 3–√

    hh100 = 3–√

    h = 3–√h3–√(100)

    (3–√1)h = 3–√(100)

    h = 3(100)31

    h = 3(100)31 ×3+13+1

    h = 3(100)(3+1)2

    = 50(3+3–√)

     

     9. The shadow of a tower on a level surface is 40 m long when moon’s elevation is 30° then when it is 60°.Find the height of the tower.

    XY=40 m, RXS =30°, RYS = 60°

    9

    Now, Consider rt.agl ΔRYS

    RYSY = tan 60°

             RYSY = 3–√

                                       h = 3–√ SY

    Consider rt.agl ΔRXQ

    RXSX = tan 30°

        RXXY+YS = 13

        h40+YS     = 13

        3–√h = 40 + YS

        3–√h = 40 + h3

    3h = 403–√ + h

    2h = 403–√

    h = 203–√

    ∴ The height of the tower is 203–√ m

     

    10. From the top of a tree h m high, the angle of depression of two objects, which are in the line with foot of the tree are γ and δ (δ> γ). Find the distance between the two objects.

    Let RS be the tree of height h m, X and Y are two objects such that

    RXS = γ , RYS = δ

    10

    Now, Consider rt.agl ΔYSR

    RSYS = tan δ

        RS = YS tan δ     …..(i)

    Now, Consider rt.agl ΔXSR

    RSXS = tan γ

    RSXY+YS = tan γ

             RS = (XY+YS) tan γ     …..(ii)

    htanγ = XY + RStanδ     [using (i)]

    h cot γ = XY + h cot δ

    XY = h (cot γ + cot δ)

     

    11. The angle of elevation of the top of a tower from two distinct points x and y from its foot are complementary. Prove that the height of the tower is xy−−√.

    Let h m be the height of the tower PQ. X and Y are two distinct points from its foot, such that PX = x and PY = y

    11

    Now, Consider rt.agl ΔXQP, we obtain

    PQPX = tan θ

    hx = tan θ

    h = x tan θ     …..(i)

    Consider rt.agl ΔYQP, we obtain

    PQPY = tan (90°θ)

    hy = cot θ

    h = y ×1tanθ   …..(ii)

    Multiply (I) and (ii), we get

    h2 = x tan θ ×y×1tanθ = xy

    h = xy−−√

    ∴ The height of the tower is xy−−√.

     

    12. The angle of elevation of the top of a tower from certain point at 30°. If the observer moves 10 m towards the tower, the angle of elevation of the top increases by 15° . Find the height of the tower.

    Let h m be the height of the tower RS and P, Q are two points 10 m apaet, such that

    SPR=30°andSQR=45°.

    12

    Now, Consider rt.agl ΔQRS

    SRQR = tan 45°

                      SR = QR = h m

    Now, Consider rt.agl ΔPRS

    SRPR = tan 30°

    h10+h = 13

            3–√ h = 10 + h

    (3–√1) h = 10

    h = 1031×3+13+1 = 10(3+1)2

    ∴ The height of the tower is 5(3–√+1)

     

    13. The angle of elevation of the top of a vertical tower from a point on the groung is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

    Let XY be the vertical tower of height h m.

    C and D are two points 10 m apart such that,

    13

    CD = ZY = 10 m XCZ=45°andXBY=60°

    XZ = XY – ZY = (h – 10) m

    Consider rt.agl ΔDYX

    XYDY = tan 60°

    hDY = 3–√

    DY = h3

    Consider rt.agl ΔCZX

    XZCZ = tan 45°

    XZ = CZ

    H – 10 = h3

    h (113) = 10

    h(313) = 10

    h = 10331×3+13+1

    = 10331×3+13+1

    = 5(3 + 3–√)m

     

    14. The angle of elevation of the top of a tower 27 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and the height of the other tower.

    Let us assume that PQ and RS be the two tower of height  h m and 27 m respectively.

    RQS = 60° PSQ = 30°

    14

    Consider rt.agl ΔRQS

    RSQS = tan 60°

    27QS = 3–√

    QS = 273     …..(i)

    Consider rt.agl ΔPSQ

    PQQS = tan30°

    hQS = 13

       3–√ h = QS     …..(ii)

    From (i) and (ii), we have

    3–√ h = 273

    h = 273 = 9 m

    From (ii), we get QS = 93–√ m

    ∴ The distance between two towers is 93–√ m and the height of  the other tower is 9 m.

     

    15. The lower window of a house is at a height of 3 m above the ground and its upper window is 6 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon from the ground.

    Let P and Q be the positions of two windows of a house, such that PR = 3 m, QR = 9 m. Let U be the position of balloon of height h m above the ground. UPS = 60°, UQT = 30°, US = h – 3 and UT = h – 9

    15

    Consider rt.agl ΔQTU

    UTQT = tan 30º

    h – 9 = QT ×13

    3–√ (h – 8) = QT      …..(i)

    Consider rt.agl ΔUPS

    USPS = tan 60º

    h – 3 = PS ×3–√

      h33 = PS      …..(ii)

    Also, QT = PS     …..(iii)

    From (i), (ii) and (iii), we get

    3–√ (h – 9) = h33

      3h -27 = h – 3

       3h – h = 27 – 3

       2h = 24

        h = 12

    ∴ height of the balloon above the ground is 12 m.

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    2017-09-28T13:24:17+00:00 Categories: CBSE|0 Comments
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