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

The base of a cliff is circular. From the extremities of a diameter of the base, angles of elevation of the top of the cliff are 30° and 60°. If the height of the cliff be 500 m, then the diameter of the base of the cliff is

(a) 2000 /\(\sqrt3\) m

(b) 1000 /\(\sqrt3\) m

(c) 2000 /\(\sqrt2\) m

(d) 1000\(\sqrt3\) m

1 Answer

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

(a) 2000 /\(\sqrt3\) m.

Let AB be the diameter of the circular base of the cliff. 

Let C be the top of the cliff.

Given, 

CE = 500 m , ∠CAE = 60°, ∠CBE = 30°,

AE = d1, BE = d2

In Δ AEC,

tan 60° = \(\frac{500}{d_1}\)

⇒ d1\(\frac{500}{\sqrt3}\) m

In Δ BEC,

tan 30° = \(\frac{500}{d_2}\)

⇒  d2 = 500\(\sqrt3\) m

∴ Required diameter = d1 + d2

\(\frac{500}{\sqrt3}\) +  500\(\sqrt3\)  = \(\frac{2000}{\sqrt3}\) m

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

...