Question

1. sarah and simone are walking in a walk - a - thon down a straight street that leads to the finish line in the park. at the same time, they both notice a hot - air balloon directly over the finish line. sarah sees the angle from the ground to the balloon as 30° and simone (who is 0.25 km closer to the finish line than sarah) sees the angle from the ground to the balloon as 45°. determine the height of the balloon, to the nearest tenth of a kilometre. 5t

Explanation:

Step1: Set up equations using trigonometry

Let the height of the balloon be \(h\) km. Let the distance of Simone from the point directly below the balloon be \(x\) km.
For Simone: \(\tan45^{\circ}=\frac{h}{x}\), so \(h = x\) (since \(\tan45^{\circ}=1\)).
For Sarah: \(\tan30^{\circ}=\frac{h}{x + 0.25}\), so \(h=(x + 0.25)\tan30^{\circ}\).

Step2: Substitute \(x = h\) into the second equation

Substitute \(x = h\) into \(h=(x + 0.25)\tan30^{\circ}\). We get \(h=(h + 0.25)\frac{1}{\sqrt{3}}\).
Multiply both sides by \(\sqrt{3}\): \(\sqrt{3}h=h + 0.25\).
Rearrange: \(\sqrt{3}h-h=0.25\), \(h(\sqrt{3}- 1)=0.25\).

Step3: Solve for \(h\)

\(h=\frac{0.25}{\sqrt{3}-1}\). Rationalize the denominator:
Multiply numerator and denominator by \(\sqrt{3}+1\).
\(h=\frac{0.25(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\).
Since \((a - b)(a + b)=a^{2}-b^{2}\), here \(a=\sqrt{3}\), \(b = 1\), so \((\sqrt{3}-1)(\sqrt{3}+1)=3 - 1=2\).
\(h=\frac{0.25(\sqrt{3}+1)}{2}\).
\(\sqrt{3}\approx1.732\), then \(h=\frac{0.25(1.732 + 1)}{2}=\frac{0.25\times2.732}{2}=\frac{0.683}{2}=0.3415\approx0.3\) (to the nearest tenth).

Answer:

The height of the balloon is approximately \(0.3\) km.