In a - ABC- a A - b B - c C where a- b and c are the sides of a triangle is equal to

The correct option is D 2(R+r)1) We know that a / 2sinA = b / 2sinB = c / 2sinC =R where, A,B,C are interior angles of ABC and a,b,c ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Construction of Incircle

In a Δ ABC,...

Question

In a $Δ A B C, (a cot A + b cot B + c cot C)$ (where $a$ , $b$ and $c$ are the sides of a triangle) is equal to

2R +r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R+2r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R+r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2(R+r)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 2(R+r)
1) We know that $\frac{a}{2 s i n A} = \frac{b}{2 s i n B} = \frac{c}{2 s i n C} = R$

where, $A, B, C$ are interior angles of $△ A B C$ and $a, b, c$ are sides of $△ A B C$
$R$ = circumradius of triangle

$∴ \frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = 2 R$

Given, $a . c o t A + b . c o t B + c . c o t C$

$= a \frac{c o s A}{s i n A} + b \frac{c o s B}{s i n B} + c \frac{c o s C}{s i n C}$

$= 2 R c o s A + 2 R c o s B + 2 R c o s C$ from (1)

$= 2 R (c o s A + c o s B + c o s C)$

We know that, $c o s A + c o s B = 2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})$
Thus, given expression becomes,

$2 R [2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2}) + c o s C]$

$= 2 R [2 c o s (\frac{π - C}{2}) c o s (\frac{A - B}{2}) + (1 - 2 {s i n}^{2} \frac{C}{2})]$
$(∵ A + B + C = π)$

$2 R [2 c o s (\frac{π}{2} - \frac{C}{2}) c o s (\frac{A - B}{2}) + (1 - 2 {s i n}^{2} \frac{C}{2})]$

$2 R [2 s i n (\frac{C}{2}) c o s (\frac{A - B}{2}) + 1 - 2 {s i n}^{2} \frac{C}{2}]$

$= 2 R [1 + 2 s i n (\frac{C}{2}) (c o s (\frac{A - B}{2}) - s i n (\frac{C}{2}))]$

$= 2 R [1 + 2 s i n (\frac{C}{2}) (c o s (\frac{A - B}{2}) - s i n (\frac{π - (A + B)}{2}))]$

$= 2 R [1 + 2 s i n (\frac{C}{2}) (c o s (\frac{A - B}{2}) - s i n (\frac{π}{2} - \frac{A + B}{2}))]$

$= 2 R [1 + 2 s i n (\frac{C}{2}) (c o s (\frac{A - B}{2}) - c o s (\frac{A + B}{2}))]$

$= 2 R [1 + 2 s i n (\frac{C}{2}) (c o s (\frac{A}{2} - \frac{B}{2}) - c o s (\frac{A}{2} + \frac{B}{2}))]$

We know that, $c o s A - c o s B = 2 s i n (\frac{A + B}{2}) s i n (\frac{B - A}{2})$
Thus, given expression becomes,

$= 2 R [1 + 2 s i n (\frac{C}{2}) (2 s i n [\frac{\frac{A}{2} - \frac{B}{2} + \frac{A}{2} + \frac{B}{2}}{2}] s i n [\frac{\frac{A}{2} + \frac{B}{2} - (\frac{A}{2} - \frac{B}{2})}{2}])]$

$= 2 R [1 + 2 s i n (\frac{C}{2}) (2 s i n \frac{A}{2} s i n \frac{B}{2})]$

$= 2 R [1 + 4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}]$

But, $4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} = \frac{r}{R}$

$= 2 R [1 + \frac{r}{R}]$

$= 2 R [\frac{R + r}{R}]$

$= 2 [R + r]$

Suggest Corrections

Similar questions

Q. Find out phenotypic categories in following cross :
A - AaBbCC

\times

aa BBCc
B - Gg Rr Tt

\times

gg rr TT
C - aabbCC

\times

AA BB cc

$A (\to a), B (\to b), C (\to c)$ are the vertices of a triangle ABC and $R (\to r)$ is any point in the plane of triangle ABC, then $\to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a)$ is always equal to

∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a x - y & y - z & z - x p - q & q - r & r - p \end{matrix} ∣ ∣ ∣ ∣

Q. If

a, b, c

are sides of the

△ A B C

such that

{(1 + \frac{b - c}{a})}^{a} \cdot {(1 + \frac{c - a}{b})}^{b} \cdot {(1 + \frac{a - b}{c})}^{c} \geq 1

, then triangle

△ A B C

must be

∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣

is equal to

Join BYJU'S Learning Program

In a ΔABC,(acotA+bcotB+ccotC) (where a,b and c are the sides of a triangle) is equal to

In a $Δ A B C, (a cot A + b cot B + c cot C)$ (where $a$ , $b$ and $c$ are the sides of a triangle) is equal to