Trigonometry & Inverse Trigonometry
Heights And Distances
nta_abhyas_2025
Grade 12
Question:
The angular depressions of the top and the foot of a tower, as seen from the top of a second tower which is $150$ meters high and standing on the same level as the first, are $13°$ and $\tan^{-1}\left(\frac{5}{6}\right)$ respectively. If the distance between their tops is $d$, then
Step-by-Step Solution
Key Concept: Use tangent ratios from two observation points to set up equations relating height and horizontal distances, then solve by equating height expressions.
We have a rectangle $CBED$ with $CE = 150$ m. Using the tangent ratios: $\tan\alpha = \frac{5}{2}$ and $\tan\beta = \frac{3}{5}$. From $\tan\alpha = \frac{h}{x}$, we get $h = \frac{5x}{2}$. From $\tan\beta = \frac{h}{150-x}$, we get $h = \frac{3(150-x)}{5}$. Equating these: $\frac{5x}{2} = \frac{3(150-x)}{5}$, which gives $25x = 6(150-x)$, so $31x = 900$, thus $x = \frac{900}{31} \approx 29$. Then $150 - x = 70$. Using the Pythagorean theorem: $AE = 80$ units, $CE = 60$ units, and $AC = 100$ units.
Correct Answer: 5