Question

43⁰ 1200 m 53⁰ H 1500 m 30 m 40 m 8. At a ski resort, the highest ski run, Hattie's Haven, can only be accessed by taking two lifts. The first lift leaves the lodge area and travels 1200 m at an inclination of 43° to a transfer point. From the transfer point, the new lift travels 1500 m at an inclination of 53° to the top of Hattie's Haven ski run. The resort is undergoing renovations and is planning on creating a new ski run called Giffin's Gallop. This run would be 30 m to the right of the top of Hattie's Haven and 40 m higher. A new lift must be installed that goes directly from the lodge area to the top of Giffin's Gallop. Determine the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop. (Thinking and Inquiry)​

Answer 1

Answer: We can use trigonometry to solve this problem. Let's call the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop "θ". Then we can break down the lift into two parts: the horizontal distance from the lodge area to the top of Giffin's Gallop, and the vertical distance from the lodge area to the top of Giffin's Gallop.

Horizontal distance:

The horizontal distance from the lodge area to the top of Giffin's Haven is 1200 m * cos(43°) + 1500 m * cos(53°) + 30 m = 1759.29 m

The horizontal distance from the top of Giffin's Haven to the top of Giffin's Gallop is 30 m.

So, the total horizontal distance from the lodge area to the top of Giffin's Gallop is 1759.29 m + 30 m = 1789.29 m

Vertical distance:

The vertical distance from the lodge area to the top of Giffin's Haven is 1200 m * sin(43°) + 1500 m * sin(53°) + 40 m = 1174.70 m

The vertical distance from the top of Giffin's Haven to the top of Giffin's Gallop is 40 m.

So, the total vertical distance from the lodge area to the top of Giffin's Gallop is 1174.70 m + 40 m = 1214.70 m

Finally, using the tangent function, we can find the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop:

θ = tan^-1(vertical distance / horizontal distance) = tan^-1(1214.70 m / 1789.29 m) = tan^-1(0.67967)

θ = 37.90° (approximately)

So, the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop is approximately 37.90°.

Explanation: