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Definite Integrals

ABC is a righ...

Question

$△ A B C$ is a right angled triangle. $O$ is the circumcenter of the $△ A B C$ and $R_{1}, R_{2}, R_{3}$ are the radii of the circumcircles of the triangle $O B A, O C A$ and $O A B$ respectively, then $\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}}$ is equal to

A

\frac{a b c}{R}

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B

\frac{a b c}{R^{3}}

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C

\frac{a b c}{R^{4}}

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D

None of these

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Solution

The correct option is A $\frac{a b c}{R^{3}}$
$R_{1} = \frac{B C}{2 sin (∠ B O C)} = \frac{a}{2 sin 2 A} \Rightarrow \frac{a}{R_{1}} = 2 sin 2 A$

Similarly, $\frac{b}{R_{2}} = 2 sin 2 B$ and $\frac{c}{R_{3}} = 2 sin 2 C$

So, $\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} = 2 (sin 2 A + sin 2 B + sin 2 C)$

$= 2.4 sin A sin B sin C [∵ A + B + C = π]$

$= (2 sin A) (2 sin B) (2 sin C) = (\frac{a}{R}) (\frac{b}{R}) (\frac{c}{C}) = \frac{a b c}{R^{3}}$

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Similar questions

Q. O is the circumcenter of

Δ A B C

R_{1}, R_{2}, R_{3}

are respectively the radii of the circumcircles of the triangles OBA,OCA and OAB

\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} = \frac{a b c}{R^{3}}

Q. O is the circumcentre of

△ A B C

R_{1}, R_{2}, R_{3}

are respectively, the radii of the circumcircles of the triangles OBC, OCA, and OAB. Prove that

\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} = \frac{a b c}{R^{3}}

O

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R_{1}, R_{2}, R_{3}

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respectively, then

\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}}

Q. O is the circumcenter of

Δ ABC and R_{1}, R_{2} and R_{3}

are the radii of the circumcircles of the triangles OBC, OCA and OAB. Then

\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} =

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