The angle of elevation of a cloud from a point $x$ $m$ above a lake is $\theta$ and the angle of depression of its reflection in the lake is ${45}^{o}$. The height of the cloud is
- $x\tan{({45}^{o}-\theta)}$
- $x\tan{({45}^{o}+\theta)}$
- $\cfrac{1}{x}\cot{({45}^{o}-\theta)}$
- $\cfrac{1}{x}\cot{({45}^{o}+\theta)}$
Let h be the height of the cloud. The distance from the point to the cloud is h-x, and the distance to the reflection is h+x. Using trigonometry, tan(theta) = (h-x)/d and tan(45) = (h+x)/d. Solving for h gives h = x * tan(45+theta).
Let the point of observation be x meters above the lake, the horizontal distance to the point directly below the cloud be d, and the cloud's height above the lake be H. The angle of elevation to the cloud is theta, so tan(theta) = (H - x) / d. The reflection of the cloud in the lake is H meters below the surface, making its vertical distance from the observation point H + x. The angle of depression to the reflection is 45 degrees, so tan(45 degrees) = (H + x) / d, which means d = H + x. Substituting d into the first equation gives tan(theta) = (H - x) / (H + x). Solving for H, we cross-multiply to get H times tan(theta) plus x times tan(theta) equals H minus x. Grouping H on one side yields H minus H times tan(theta) equals x times tan(theta) plus x. Factoring out H and x gives H times (1 minus tan(theta)) equals x times (1 plus tan(theta)). Using the tangent addition formula, the ratio (1 plus tan(theta)) divided by (1 minus tan(theta)) equals tan(45 degrees plus theta). Therefore, H equals x multiplied by tan(45 degrees plus theta).