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
128 views
in Trigonometry by (67.7k points)
closed by
A variable triangle `A B C` is circumscribed about a fixed circle of unit radius. Side `B C` always touches the circle at D and has fixed direction. If B and C vary in such a way that (BD) (CD)=2, then locus of vertex A will be a straight line. parallel to side BC perpendicular to side BC making an angle `(pi/6)` with BC making an angle `sin^(-1)(2/3)` with `B C`

1 Answer

0 votes
by (68.9k points)
selected by
 
Best answer
Here, `BD = s-b`
`CD = s-c`
`BD*CD = (s-b)(s-c)`
`=>2 = (s-b)(s-c)`
`=>2s(s-a) = s(s-a)(s-b)(s-c)`
`=>2s(s-a) = Delta^2`
`=>Delta^2/s^2 = (2(s-a))/s`
`=>r^2 = 2-(2a)/s`
As, `r` is constant, so `a/s` will be constant.
Let `H_a` is the distance of `A` from `BC`.
Then, `Delta = 1/2*a*H_a`
`=>Delta/s = 1/(2s)aH_a`
It is given that circle is of unit radius,
` :. 1 = a/(2s)H_a`
`=>H_a = (2s)/a`
It means `H_a` will be constant as `a/s` is a constant.
So, locus of vertex `A` will be a straight line parallel to `BC`.
So, option `(a)` is the correct option.

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

...