Trigonometry
Heights and Distances
MMTS_Full_Test_05
Grade 12
Question:
The upper four-fifth portion of a vertical tower subtends an angle $\tan^{-1}\frac{8}{21}$ at a point $A$ in the horizontal plane through its foot and at a distance 50 m from the foot. If the angle subtended by the lower one fifth of tower at point $A$ is $\beta$, then the height of the tower can be
Step-by-Step Solution
Key Concept: Let height $h$. Lower $\frac{h}{5}$, upper $\frac{4h}{5}$. $\tan\beta=\frac{h/5}{50}$ and $\tan(\alpha+\beta)-\tan\beta=\tan^{-1}(8/21)$... where $\alpha$ is angle to top.
Let $\tan\beta=\frac{h/5}{50}=\frac{h}{250}$, $\tan(\alpha+\beta)=\frac{h}{50}$ (angle to full tower). $\tan\alpha=\tan((\alpha+\beta)-\beta)=\frac{h/50-h/250}{1+h^2/12500}=\frac{4h/250}{1+h^2/12500}=\frac{8}{21}$. Cross multiply: $21\cdot 4h\cdot 12500 = 8\cdot 250(12500+h^2)$. $21h\cdot 50=8(12500+h^2)/10$... Solving: $h^2-\frac{105\cdot 50}{8}h+12500=0$... numerically $h=25$ satisfies.
Correct Answer: A