Question

1. two ropes are attached to the bottom of a hot air balloon. the ropes are also attached to the ground on opposite sides of the balloon. the first rope is 20 m long and makes an angle of 30° with the ground. the second rope is 25 m long. determine the distance on the ground between the points where the two ropes are attached. a - 5t - 5

Explanation:

Step1: Find the length of MH

In right - triangle with the first rope, we know the length of the rope (hypotenuse) is 20 m and the angle with the ground is 30°. Using the cosine function $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Let the adjacent side to the 30° angle be $MH$. So, $\cos30^{\circ}=\frac{MH}{20}$. Then $MH = 20\times\cos30^{\circ}=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\text{ m}$.

Step2: Find the length of NH

In right - triangle with the second rope, we can use the Pythagorean theorem. Let the height of the balloon above the ground be $h$. From the first right - triangle, $h = 20\times\sin30^{\circ}=10\text{ m}$. In the second right - triangle with hypotenuse 25 m and height $h = 10\text{ m}$, using $a^{2}+b^{2}=c^{2}$, where $c = 25$ and $a = 10$. Let the base be $NH$. Then $NH=\sqrt{25^{2}-10^{2}}=\sqrt{(25 + 10)(25 - 10)}=\sqrt{35\times15}=\sqrt{525}=5\sqrt{21}\text{ m}$.

Step3: Find the distance between M and N

The distance between the two points where the ropes are attached on the ground is $MN=MH + NH$. So $MN=10\sqrt{3}+5\sqrt{21}\approx10\times1.732+5\times4.583=17.32 + 22.915=40.235\text{ m}$.

Answer:

$10\sqrt{3}+5\sqrt{21}\text{ m}\approx40.24\text{ m}$