A weather balloon that is rising vertically is being observed from a point on the ground $300 \mathrm{ft}$ from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observer's line of sight is $45^{\circ}$ and is increasing at $1^{\circ}$ per second?
Added by Brian P.
Step 1
Let's denote the angle between the ground and the observer's line of sight as $\theta$. We are given that $\theta = 45^{\circ}$ and $\frac{d\theta}{dt} = 1^{\circ}/s$. Show more…
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