Two trees stand on opposite sides of a straight road. The distance between the two trees is 400 metres. A point P lies on the road between them. The angles of depression of point P from the tops of the two trees are 45° and 60°. If the height of the tree that makes the 45° angle of depression is 200 metres, what is the height (in metres) of the other tree?

Introduction / Context:
This problem is a classic height and distance question from trigonometry. We are given angles of depression from the tops of two trees to a point on a road between them, and the height of one tree. Using basic trigonometric ratios, we can find the height of the second tree.

Given Data / Assumptions:

• Two trees stand on opposite sides of a straight road.
• Distance between the trees along the road = 400 metres.
• A point P lies on the road between the trees.
• Angle of depression to P from the top of the first tree = 45°.
• Height of this first tree = 200 metres.
• Angle of depression to P from the top of the second tree = 60°.
• The road is assumed to be horizontal and the trees vertical.

Concept / Approach:
Angles of depression from the top of a vertical object to a point on the ground correspond to angles of elevation from that ground point back to the top. In a right triangle, tan(θ) = opposite / adjacent, where the opposite side is the height and the adjacent side is the horizontal distance from the point on the ground to the foot of the tree. We will set up two such right triangles, use tangent values for 45° and 60°, and use the fact that the point lies between the two trees so that the two horizontal distances add up to 400 metres.

Step-by-Step Solution:
Let the trees be at points A and B, with A on the left and B on the right, and let P be the point on the road between them. Let AP be the horizontal distance from the foot of the 200 metre tree to point P. For the first tree, tan 45° = height / AP = 200 / AP. Since tan 45° = 1, we get 1 = 200 / AP, so AP = 200 metres. The total distance between the trees is AB = 400 metres, so the remaining distance BP = AB - AP = 400 - 200 = 200 metres. Let the height of the second tree be h metres. For the second tree, tan 60° = h / BP = h / 200. We know tan 60° = √3, so √3 = h / 200, which gives h = 200√3 metres.

Verification / Alternative check:
We can quickly check reasonableness. The point P is equally distant horizontally from both trees (200 metres each side). Since the angle of depression from the taller tree is 60°, which is larger than 45°, the taller tree must have greater height. Using h = 200√3 gives h approximately equal to 200 × 1.732 = 346.4 metres, which is indeed taller than 200 metres and consistent with a larger angle.

Why Other Options Are Wrong:
200: This would mean both trees have the same height, which cannot produce different angles of depression from a symmetric position on the road.
100√3: This is less than 200, so it would give a smaller angle of depression than 45°, not 60°.
250: This height is larger than 200 but does not satisfy tan 60° = height / 200, so it does not match the trigonometric condition.

Common Pitfalls:
Students often mix up angles of elevation and depression or forget that the point lies between the trees, so the two horizontal distances must sum to 400 metres. Another common mistake is to use sine instead of tangent, or to incorrectly take the distance between trees as one of the legs of the right triangle without splitting it at the point P.

Final Answer:
Thus, the height of the other tree is 200√3 metres.