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8. a lighthouse operator at 25 m above sea level sights a sailboat. the…

Question

  1. a lighthouse operator at 25 m above sea level sights a sailboat. the angle of depression of the sighting is 10°. how far is the boat from the base of the lighthouse?

Explanation:

Step1: Identify the trigonometric relationship

We have a right - triangle where the height of the lighthouse (opposite side) is \(h = 25\) m and the angle of depression \(\theta=10^{\circ}\). The angle of depression is equal to the angle of elevation from the boat to the top of the lighthouse. Let \(x\) be the distance from the boat to the base of the lighthouse (adjacent side). We use the tangent function \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
So, \(\tan(10^{\circ})=\frac{25}{x}\)

Step2: Solve for \(x\)

We can rewrite the equation as \(x = \frac{25}{\tan(10^{\circ})}\).
We know that \(\tan(10^{\circ})\approx0.1763\)
Then \(x=\frac{25}{0.1763}\approx141.8\)

Answer:

The boat is approximately \(141.8\) m from the base of the lighthouse.