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  1. Asked: In:

    A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

    jai kohli

    Let K be the Kite and P is the point where the string is tied on the ground. In ΔKLP, KL/KP = sin60° ⇒ 60/KP = √3/2 ⇒ AB = 120/√3 = 120√3/3 = 40√3 Hence, the lenght of the string is 40√3 m.

    Let K be the Kite and P is the point where the string is tied on the ground.
    In ΔKLP,
    KL/KP = sin60°
    ⇒ 60/KP = √3/2
    ⇒ AB = 120/√3 = 120√3/3 = 40√3
    Hence, the lenght of the string is 40√3 m.

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  2. Asked: In:

    A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

    jai kohli

    Let PQ is building and AS is a boy 1.5 m tall. He moves from S to towards building. Here, PQ = 30m Therefore, PR = PQ - RQ = (30 - 1.5) m = 28.5 m = 57/2 m In ΔPAR, PR/AR = tan 30° ⇒ (57/2)/AR = 1/√3 ⇒ AR = 57√3/2 In ΔPRB, PR/BR = tan 60° ⇒ (57/2)/BR = √3 ⇒ BR = 57/2√3 = 57√3/6 = 19√3/2 Distance walRead more

    Let PQ is building and AS is a boy 1.5 m tall. He moves from S to towards building.
    Here, PQ = 30m
    Therefore, PR = PQ – RQ = (30 – 1.5) m = 28.5 m = 57/2 m
    In ΔPAR,
    PR/AR = tan 30° ⇒ (57/2)/AR = 1/√3 ⇒ AR = 57√3/2
    In ΔPRB,
    PR/BR = tan 60° ⇒ (57/2)/BR = √3 ⇒ BR = 57/2√3 = 57√3/6 = 19√3/2
    Distance walked towards the building
    = ST = AB = AR – BR = 58√3/2 – 19√3/2 = 38√3/2 = 19√3
    Hence, the distance walked towards the building is 19√3 m.

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  3. Asked: In:

    A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

    jai kohli

    In the given figure, AB is pole and AC is rope. In ΔABC, AB/AC = sin30⁰ ⇒ AB/20 = 1/2 ⇒ AB = 20/2 = 10 Hence, the height of the pole is 10 m.

    In the given figure, AB is pole and AC is rope.
    In ΔABC,
    AB/AC = sin30⁰ ⇒ AB/20 = 1/2 ⇒ AB = 20/2 = 10
    Hence, the height of the pole is 10 m.

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  4. Asked: In:

    A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

    jai kohli

    Let AC is the tree which break at B and the broken part AB become A'B. In ΔA'BC, BC/A'C = ten 30⁰ ⇒ BC/8 = 1/√3 ⇒ BC = 8/√3 and A'C/A'B = cos 30⁰ ⇒ 8/A'B = √3/2 ⇒ A'B = 16/√3 Height of the tree = AC = AB + BC = A'B + BC = 16/√3 + 8/√3 = 24/√3 = 24√3/3 = 8√3 Hence, the height of the tree 8/√3m.

    Let AC is the tree which break at B and the broken part AB become A’B.
    In ΔA’BC,
    BC/A’C = ten 30⁰
    ⇒ BC/8 = 1/√3
    ⇒ BC = 8/√3
    and A’C/A’B = cos 30⁰
    ⇒ 8/A’B = √3/2 ⇒ A’B = 16/√3
    Height of the tree
    = AC = AB + BC = A’B + BC
    = 16/√3 + 8/√3 = 24/√3 = 24√3/3 = 8√3
    Hence, the height of the tree 8/√3m.

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  6. Asked: In:

    A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

    jai kohli

    Let AC be the slide for the children below the age of 5 years. The height of this slide is 1.5 m inclined at 30°. In ΔABC, AB/AC = sin 30° ⇒ 1.5/AC = 1/2 ⇒ AC = 3 Let AC be the slide for the childern above 5 years which is 3 m high and inclined at 60° with the horizontal. In ΔABC, AB/AC = sin 60° ⇒Read more

    Let AC be the slide for the children below the age of 5 years. The height of this slide is 1.5 m inclined at 30°.
    In ΔABC,
    AB/AC = sin 30°
    ⇒ 1.5/AC = 1/2
    ⇒ AC = 3
    Let AC be the slide for the childern above 5 years which is 3 m high and inclined at 60° with the horizontal.
    In ΔABC,
    AB/AC = sin 60° ⇒ 3/AC = √3/2
    ⇒ AC = 6/√3 = 6√3/3 = 2√3
    Hence, the height of the slides are 3m and 2√3.

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