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.
For Simone: \(\tan45^{\circ}=\frac{h}{x}\), so \(h = x\) (since \(\tan45^{\circ}=1\)).
For Sarah: \(\tan30^{\circ}=\frac{h}{x + 0.25}\), so \(h=\frac{1}{\sqrt{3}}(x + 0.25)\) (since \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\)).

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

Substitute \(h=x\) into \(h=\frac{1}{\sqrt{3}}(x + 0.25)\):
\(x=\frac{1}{\sqrt{3}}(x + 0.25)\)
Multiply both sides by \(\sqrt{3}\): \(\sqrt{3}x=x + 0.25\)
Rearrange: \(\sqrt{3}x-x=0.25\)
Factor out \(x\): \(x(\sqrt{3}-1)=0.25\)
Solve for \(x\): \(x=\frac{0.25}{\sqrt{3}-1}\)
Rationalize the denominator: \(x=\frac{0.25(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\)
Since \((a - b)(a + b)=a^{2}-b^{2}\), here \(a=\sqrt{3}\), \(b = 1\), so \((\sqrt{3}-1)(\sqrt{3}+1)=3 - 1=2\)
\(x=\frac{0.25(\sqrt{3}+1)}{2}\)
Calculate \(\sqrt{3}\approx1.732\): \(x=\frac{0.25(1.732 + 1)}{2}=\frac{0.25\times2.732}{2}=0.3415\)

Answer:

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