Question

An airplane climbing at a constant angle relative to the ground has reached an altitude of 4 km.At this point, the plane has traveled a horizontal distance of 14 km.What is the angle at which the airplane is climbing?

Answer 1

Answer:

$θ = tan⁻¹(\frac{2}{7})$

Step-by-step explanation:

Let θ be the angle at which the aeroplane is climbing.

We have,

Δx = 14 Km

Δy = 4 Km

$so, tan(θ) = \frac{Δy}{Δx}$

$or\:\:tan(θ) = \frac{4}{14}$

$or\:\:tan(θ) = \frac{2}{7}$

$or\:\:θ = tan⁻¹(\frac{2}{7})$