Question

I hot-air ballon is held at a constant altitude by two ropes that are anchored to the ground. One rope is 110 feet long and makes an angle of 55° with the ground. The other rope is 105 feet long. What is the distance between the points on the ground at which the two ropes are anchored? (Enter your answer as a comma separated list. And your answer to the nearest whole number.)

Answer 1

so hmm check the picture below

the first part, on the left is the guywire one

so let's use the law of sines


`\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\ -----------------------------\\\\ \cfrac{sin(56^o)}{43}=\cfrac{sin(B)}{51}\implies \cfrac{51\cdot sin(56^o)}{43}=sin(B) \\\\\\ sin^(-1)\left[ \cfrac{51\cdot sin(56^o)}{43} \right]=\measuredangle B\implies 79.5\approx B`

now, if the angle B is 79.5 and the other angle is 56, then A is the slack from 180, or 180 - 79.5 - 56, or 44.5

so, let's see what's the opposite side of A, or "x"


`\bf \cfrac{sin(56^o)}{43}=\cfrac{sin(A)}{x}\implies \cfrac{sin(56^o)}{43}=\cfrac{sin(44.5^o)}{x}\\\\\\x=\cfrac{43\cdot sin(44.5^o)}{sin(56^o)}`

---------------------------------------------------------------------------

now, the second one, you had in the picture, the one of the balloon
that's the right-side in the picture below

again, let's use the law of sines


`\bf \cfrac{sin(55^o)}{105}=\cfrac{sin(B)}{110}\implies \cfrac{110\cdot sin(55^o)}{105}=sin(B) \\\\\\ sin^(-1)\left[ \cfrac{110\cdot sin(55^o)}{105} \right]=\measuredangle B\implies 59.1\approx B`

so if B is 59.1, angle A picks up the slack from 180, or 180 - 59.1 - 55, or 65.9

so, let's see what "x" is then
`\bf \cfrac{sin(55^o)}{105}=\cfrac{sin(A)}{x}\implies \cfrac{sin(55^o)}{105}=\cfrac{sin(65.9^o)}{x} \\\\\\ x=\cfrac{105\cdot sin(65.9^o)}{sin(55^o)}`


on both cases, since the angles are in degrees, make sure your calculator is in Degree mode