Certainly! Let's go through the problem step-by-step with a detailed explanation. --- # Problem Summary: - A helicopter is flying at a height of 150 meters. - It is moving horizontally at a speed of 125 m/s. - The helicopter is observed from a point on the ground, which is 65 meters above the level of the helicopter's initial position. - The horizontal angle of elevation to the helicopter changes at a rate of .12 radians/sec. - You are asked to find **the height of the helicopter when the angle of elevation is \( \theta = 60^\circ \)**. --- # Step 1: Understand the diagram and variables Let's define the variables: - \( H \): height of the helicopter (unknown, to be found) - \( D \): horizontal distance from the observer to the point directly below the helicopter at the instant of interest. - \( \theta \): angle of elevation from the observer to the helicopter. - \( y \): the vertical distance from the observer's eye to the helicopter, which is \( H - 65 \) meters (since observer is 65 meters above the initial level). From the diagram: - The observer's eye level is 65 meters above the ground. - The helicopter's height above the ground is \( H \). --- # Step 2: Express the relationship between variables Using the right triangle with the angle of elevation \( \theta \): \[ \tan \theta = \frac{\text{vertical distance from observer to helicopter}}{\text{horizontal distance}} \] Thus: \[ \tan \theta = \frac{H - 65}{D} \] Rearranged: \[ D = \frac{H - 65}{\tan \theta} \] --- # Step 3: Relate the rates Given: - Rate of change of \( \theta \): \( \frac{d\theta}{dt} = .12 \, \text{rad/sec} \) Differentiate both sides of the equation \( \tan \theta = \frac{H - 65}{D} \) with respect to time \( t \): \[ \frac{d}{dt} \left( \tan \theta \right) = \frac{d}{dt} \left( \frac{H - 65}{D} \right) \] Since \( H \) and 65 are constants (assuming the helicopter's height is constant during the observation): \[ \sec^2 \theta \cdot \frac{d\theta}{dt} = - \frac{H - 65}{D^2} \cdot \frac{dD}{dt} \] Rearranged to find \( \frac{dD}{dt} \): \[ \frac{dD}{dt} = - \frac{D^2}{H - 65} \cdot \sec^2 \theta \cdot \frac{d\theta}{dt} \] --- # Step 4: Find \( D \) at \( \theta = 60^\circ \) Substitute \( \theta = 60^\circ \): \[ \tan 60^\circ = \sqrt{3} \approx 1.732 \] \[ D = \frac{H - 65}{\tan 60^\circ} = \frac{H - 65}{\sqrt{3}} \] --- # Step 5: Use the helicopter's horizontal speed The helicopter moves horizontally at \( 125 \, \text{m/sec} \). The horizontal rate of change of \( D \) is: \[ \frac{dD}{dt} = -125 \, \text{m/sec} \] (negative because as the helicopter moves away, \( D \) increases, but for the purposes of the rate of change, the sign is consistent with the direction). --- # Step 6: Calculate \( D \) at \( \theta = 60^\circ \) Now, plug into the earlier derivative formula: \[ -125 = - \frac{D^2}{H - 65} \cdot \sec^2 60^\circ \cdot .12 \] Recall: \[ \sec 60^\circ = \frac{1}{\cos 60^\circ} = 2 \] \[ \sec^2 60^\circ = 4 \] Therefore: \[ -125 = - \frac{D^2}{H - 65} \times 4 \times .12 \] Simplify: \[ -125 = - \frac{D^2}{H - 65} \times .48 \] Multiply both sides by -1: \[ 125 = \frac{D^2}{H - 65} \times .48 \] Express \( D \) in terms of \( H \): \[ D = \frac{H - 65}{\sqrt{3}} \] Plug into the equation: \[ 125 = \frac{\left( \frac{H - 65}{\sqrt{3}} \right)^2}{H - 65} \times .48 \] Simplify numerator: \[ \left( \frac{H - 65}{\sqrt{3}} \right)^2 = \frac{(H - 65)^2}{3} \] So: \[ 125 = \frac{\frac{(H - 65)^2}{3}}{H - 65} \times .48 \] Simplify the fraction: \[ \frac{(H - 65)^2}{3(H - 65)} = \frac{H - 65}{3} \] So the equation becomes: \[ 125 = \frac{H - 65}{3} \times .48 \] Multiply both sides by 3: \[ 375 = (H - 65) \times .48 \] Divide both sides by .48: \[ \frac{375}{.48} = H - 65 \] Calculate: \[ H - 65 = 781.25 \] Finally, solve for \( H \): \[ H = 781.25 + 65 = \boxed{846.25 \text{ meters}} \] --- # **Final Answer:** \[ \boxed{ \text{The height of the helicopter when the angle of elevation is } 60^\circ \text{ is approximately } \mathbf{846.25 \text{ meters}} } \] --- Let me know if you'd like any further clarification!