3. sarah and simone are walking in a walk - a - thon down a straight street that leads to the finish line in…

3. 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

3. 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

Answer

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