Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
59 views
in Trigonometry by (48.0k points)
closed by

The angle of elevation of the top of a TV tower from three points A, B, C in a straight line (in a horizontal plane) through the foot of the tower are α, 2α, 3α respectively. If AB = a, the height of the tower is

(a) a sin α

(b) a sin 2α

(c) a sin 3α

(d) a sin (α2)

1 Answer

+1 vote
by (49.1k points)
selected by
 
Best answer

Correct option is (b) a sin 2α

Let EF be the TV tower such that

∠EAF = α, ∠EBF = 2α,

∠ECF = 3α and AB = a

In \(\Delta\)EBA,

ext. ∠EBF = ∠BEA + ∠BAE 

⇒ 2α = ∠BEA + α 

⇒ ∠BEA = α

\(\Delta\)BEA is isosceles 

⇒ BE = a.

In rt. \(\Delta\)EBF, sin 2α = \(\frac{EF}{EB}\)

⇒ EF = EB sin 2α = a sin 2α

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...