A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to a point on the ground which is at a certain distance from the foot of the pole. The height of the pole, if the angle made by the rope with the ground level is $30^{\circ}$ is ___ m.

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Solution

The figure above represents the given situation.
In the given figure,
$s i n 30^{\circ} = \frac{A B}{A C} \Rightarrow A B = s i n 30^{\circ} \times A C = \frac{A C}{2} = \frac{30}{2} = 15 m$ .

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A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. The height of the pole is ___ m , if the angle made by the rope with the ground level is $30^{\circ}$ .

Q. A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to a point on the ground which is at a certain distance from the foot of the pole. If the angle made by the rope with the ground level is

30^{\circ}

, the height of the pole is ___ m.

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A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to a point on the ground which is at a certain distance from the foot of the pole. The height of the pole, if the angle made by the rope with the ground level is 30∘ is ___ m.

The figure above represents the given situation. In the given figure, sin 30∘=ABAC⇒AB=sin 30∘×AC=AC2=302=15 m.

A circus artist is climbing a 30 m long rope, which is tightly stretched and tied from the top of a vertical pole to a point on the ground which is at a certain distance from the foot of the pole. The height of the pole, if the angle made by the rope with the ground level is $30^{\circ}$ is ___ m.

The figure above represents the given situation.
In the given figure,
$s i n 30^{\circ} = \frac{A B}{A C} \Rightarrow A B = s i n 30^{\circ} \times A C = \frac{A C}{2} = \frac{30}{2} = 15 m$ .