Fewpal
+3 votes
584 views
in Mathematics by (64.1k points)

The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.

1 Answer

+2 votes
by (127k points)
selected by
 
Best answer

Answer

image

Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
A/q,
In  right ΔABC,
tan x = AB/BC
⇒ tan = AB/4
⇒ AB = 4 tan x ... (i)
also,
In  right ΔABD,
tan (90°-x) = AB/BD
⇒ cot = AB/9
⇒ AB = 9 cot  ... (ii)
Multiplying  eqn (i) and (ii)
AB2 = 9 cot × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

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

...