A tent 10 m high is tied with a taut rope from the ground to the top of the tent making an angle of elevation of $30^{\circ}$ with the ground. The length of the rope is ___ m

Open in App

Solution

Let the length of the string be l .

sin $30^{\circ}$ = $\frac{10}{I}$

$\Rightarrow$ $l$ = $\frac{10}{1 / 2}$ = 20m

Suggest Corrections

Similar questions

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, 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 the ground. Find the distance of the pole to the peg in the ground, if the angle made by the rope with the ground level is

30^{\circ}

Q. A rope of length

5

metres is tightly tied with one end at the top of a vertical pole and other end to the horizontal ground. If the rope makes an angle

30^{0}

to the horizontal, then the height of the pole is

A circus artist is climbing a 20m long rope which is tightly stretched and tied from the top of a vertical pole to the ground. If the rope makes an agle of $30^{\circ}$ with the ground, then the height of the pole is ___ metres.

Join BYJU'S Learning Program

A tent 10 m high is tied with a taut rope from the ground to the top of the tent making an angle of elevation of 30∘ with the ground. The length of the rope is ___ m

Let the length of the string be l . sin30∘ = 10I ⇒ l = 101/2 = 20m

A tent 10 m high is tied with a taut rope from the ground to the top of the tent making an angle of elevation of $30^{\circ}$ with the ground. The length of the rope is ___ m

Let the length of the string be l .

sin $30^{\circ}$ = $\frac{10}{I}$

$\Rightarrow$ $l$ = $\frac{10}{1 / 2}$ = 20m