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 and the distance from Simone to the point directly below the balloon be \( x \) km.
From Simone's perspective (\(45^{\circ}\) angle), \(\tan45^{\circ}=\frac{h}{x}\), so \(h = x\).
From Sarah's perspective (\(30^{\circ}\) angle), \(\tan30^{\circ}=\frac{h}{x + 0.25}\). Since \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), we have \(\frac{h}{h + 0.25}=\frac{1}{\sqrt{3}}\).

Step2: Solve the equation for \(h\)

Cross - multiply the equation \(\frac{h}{h + 0.25}=\frac{1}{\sqrt{3}}\) to get \(\sqrt{3}h=h + 0.25\).
Rearrange terms: \(\sqrt{3}h−h = 0.25\), factor out \(h\): \(h(\sqrt{3}-1)=0.25\).
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 \(\sqrt{3}\approx1.732\), so \(h=\frac{0.25(1.732 + 1)}{2}=\frac{0.25\times2.732}{2}=0.3415\approx0.3\) (to the nearest tenth).

Answer:

\(0.3\) km