Question

mcr3u
trigonometry assignment t /29

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 (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 \(h=x\), we substitute \(x\) with \(h\) in the second - equation. So \(\tan30^{\circ}=\frac{h}{h + 0.25}\).

Step2: Solve the equation for \(h\)

We know that \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\approx0.577\). The equation \(\frac{1}{\sqrt{3}}=\frac{h}{h + 0.25}\) can be cross - multiplied.
\(h+0.25=\sqrt{3}h\).
\(0.25=\sqrt{3}h−h\).
\(0.25 = h(\sqrt{3}-1)\).
\(h=\frac{0.25}{\sqrt{3}-1}\).
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 \((\sqrt{3}-1)(\sqrt{3}+1)=3 - 1=2\).
\(h=\frac{0.25(\sqrt{3}+1)}{2}\).
\(h=\frac{0.25\times(1.732 + 1)}{2}\) (since \(\sqrt{3}\approx1.732\)).
\(h=\frac{0.25\times2.732}{2}\).
\(h=\frac{0.683}{2}=0.3415\approx0.3\) km.

Answer:

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