Question : A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height $h$ units. At a point on the plane, the angle of elevation at the bottom of the flagstaff is $\alpha$ and that of the top of the flagstaff is $\beta$. Then the height of the tower is:
Option 1: $h\tan\alpha$
Option 2: $\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$
Option 3: $\frac{h\tan\alpha}{(\tan\alpha–\tan\beta)}$
Option 4: None of these
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$
Solution :
Given: A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. The angle of elevation at the bottom of the flagstaff is $\alpha$ and that of the top of the flagstaff is $\beta$.
We know that $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
Let AB be $x$.
From the figure, it is clear that in $\triangle ADB$, $\tan\alpha =\frac{AD}{AB}$
$\tan\alpha =\frac{H}{x}$
⇒ $x =\frac{H}{\tan\alpha}$
Similarly, in $\triangle CAB$, $\tan\beta =\frac{CA}{AB}$
$\tan\beta =\frac{h+H}{x}$
⇒ $x =\frac{h+H}{\tan\beta}$
So, $\frac{h+H}{\tan\beta}=\frac{H}{\tan\alpha}$
⇒ $H\tan\beta=h\tan\alpha+H\tan\alpha$
⇒ $H (\tan\beta–\tan\alpha)=h\tan\alpha$
⇒ $H =\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$
Hence, the correct answer is $\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Trending Articles around SSC CGL
