TF is a tower with F on the ground. The angle of elevation of T from A is $x^\circ$ such that $\tan x^\circ = \frac{2}{5}$ and AF = 200m. The angle of elevation of T from a nearer point B is $y^\circ$ with BF = 80m. The value of $y^\circ$ is

### Concept & Stepwise Triangle Solving When given two points on the same line from a tower, use the point with the known angle and distance to find the tower's height first. Then, use that height to find the unknown angle for the second point. $$ \text{Height} = \text{Distance} \times \tan(\theta) $$ ### Step-by-Step Solution 1. Let the height of the tower TF be $h$. The base is F. 2. Consider the first point A: The distance AF = $200$ m, and the angle of elevation is $x^\circ$ with $\tan x^\circ = \frac{2}{5}$. 3. Using the tangent ratio in triangle TAF: $$ \tan x^\circ = \frac{h}{AF} \Rightarrow \frac{2}{5} = \frac{h}{200} $$ 4. Solve for $h$: $$ h = 200 \times \frac{2}{5} = 40 \times 2 = 80 \text{ m} $$ 5. Now consider the nearer point B: The distance BF = $80$ m, and the angle of elevation is $y^\circ$. 6. Using the tangent ratio in triangle TBF: $$ \tan y^\circ = \frac{h}{BF} $$ 7. Substitute the known values $h = 80$ and $BF = 80$: $$ \tan y^\circ = \frac{80}{80} = 1 $$ 8. The angle whose tangent is $1$ is $45^\circ$. Therefore, $y^\circ = 45^\circ$. ### Exam Strategy & Shortcut Work backwards intuitively. If $\tan x = 2/5$, and the base is $200$, the height is scaled by $40$ ($2 \times 40 = 80$). Now for point B, the base is $80$. A triangle with height $80$ and base $80$ is an isosceles right triangle, meaning the base angle is automatically $45^\circ$. ### Common Pitfall Overcomplicating the algebra or confusing the distances AF and BF, which would result in calculating the incorrect height and subsequently the wrong angle. ### Final Answer Therefore, the correct answer is **$45^\circ$**.