the angle of elevation of the top of top of a tower at a distance 150m from its foot on a horizontal plane is found to be 30 find the high

Question

the angle of elevation of the top of top of a tower at a distance 150m from its foot on a horizontal plane is found to be 30 find the hight of the tower​

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Skylar 2 months 2021-08-11T13:52:33+00:00 2 Answers 0 views 0

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    0
    2021-08-11T13:53:47+00:00

    Answer:

    Height of the tower is 51.96 m.

    Step-by-step explanation:

    Let AB be the tower and O be the point of observation.

    Then OA = 90 cm, ∠OAB = 90°.

    Let AB = h meters. Then from right Δ AOB, ∠AOB = 30°.

    We have,

    \frac{AB}{OA}=\frac{Perpendicular}{Base}=tan30^\o

    \frac{h}{90^0}=\frac{1}{\sqrt3}

    h=\frac{1}{\sqrt3}\times90^0

    h=\frac{90}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}

    h=\frac{90}{3}\times\sqrt3

    h=30\times\sqrt3

    h=30\times1.732

    ∴ The height of the tower is 51.96 m.

    0
    2021-08-11T13:54:28+00:00

    Answer:

    50

     \sqrt{3}

    Step-by-step explanation:

    The distance from the foot of the tower is 150m and the angle of elevation is 30 degrees.

    = tan30 degree=Perpendicular/Base

    = 1/root3=P/150

    = 150/root3=P

    Then, by rationalising the denominator the answer will be

    = P=50root3

    = So the height of the tower is 50root3

    Hope it helps.

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