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
2.3k views
in Trigonometry by (36.2k points)
closed by

The angle of elevation of the top of a tower from the top and bottom of a building of height a are 30° and 45° respectively. If the tower and the building stand at the same level, then the height of the tower is

(a) \(a\sqrt3+1\)

(b) \(\frac{a\sqrt3}{\sqrt3+1}\)

(c) \(\frac{a\sqrt3(\sqrt3+1)}{2}\)

(d) \(\frac{a+\sqrt3}{2(\sqrt3-1)}\)

1 Answer

+1 vote
by (33.4k points)
selected by
 
Best answer

(c) \(\frac{a\sqrt3(\sqrt3+1)}{2}\)

Let AB be the tower of height h metres (say).

Given, 

building PQ = a metres. Draw PR || BQ such that R lies on AB.

⇒ PR = BQ

Given,

∠APR = 30° and ∠AQB = 45°

RB = PQ = a

⇒ AR = (h – a)

∴ In rt. Δ ARP,

tan 30° = \(\frac{AR}{RP}=\frac{h-a}{RP}\)

⇒ \(\frac{1}{\sqrt3}=\frac{h-a}{RP}\)

⇒ RP = (h-a)\(\sqrt3\) ...(i)

In rt. Δ ABQ,

 tan 45° = \(\frac{AB}{BQ}=\frac{h}{BQ}\)

⇒ 1 = \(\frac{h}{BQ}\)

⇒ BQ = h ...(ii)

∴ RP = BQ, from (i) and (ii), we have,

h = (h - a)\(\sqrt3\) 

⇒  h(\(\sqrt3\) - 1) = a\(\sqrt3\)

⇒ h = \(\frac{a\sqrt3}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}\)

\(\frac{a\sqrt3(\sqrt3+1)}{2}\) 

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

...