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.
For Simone (angle \(45^{\circ}\)), \(\tan45^{\circ}=\frac{h}{x}\), so \(h = x\) (since \(\tan45^{\circ}=1\)).
For Sarah (angle \(30^{\circ}\)), \(\tan30^{\circ}=\frac{h}{x + 0.25}\).
Since \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), we have \(\frac{1}{\sqrt{3}}=\frac{h}{h + 0.25}\).

Step2: Solve the equation for \(h\)

Cross - multiply: \(h+0.25=\sqrt{3}h\).
Rearrange: \(0.25=\sqrt{3}h - h\).
Factor out \(h\): \(0.25=h(\sqrt{3}-1)\).
Then \(h=\frac{0.25}{\sqrt{3}-1}\).
Rationalize the denominator: \(h=\frac{0.25(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\).
Since \((a - b)(a + b)=a^{2}-b^{2}\), here \(a=\sqrt{3}\), \(b = 1\), so \((\sqrt{3}-1)(\sqrt{3}+1)=3 - 1=2\).
Then \(h=\frac{0.25(\sqrt{3}+1)}{2}\).
Calculate: \(h=\frac{0.25\times1.732+0.25}{2}=\frac{0.433 + 0.25}{2}=\frac{0.683}{2}=0.3415\approx0.3\) (to the nearest tenth).

Answer:

\(0.3\) km