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Answer 1

Given:

A triangle ABC in which sides are a, b and c.

To Prove:

$\large\bold\red{\frac{a}{ sin A } = \frac{b}{ sin B } = \frac{c}{ sin C }}$

Proof:

Let,

$\vec{AB} = \vec{c} \\ \\ \vec{BC} = \vec{a} \\ \\ \vec{CA} = \vec{b}$

So,

$\vec{a} + \vec{b} + \vec{c} = 0$

Now,

Using the Properties of Cross Product

we get,

$= > \vec{a} \times (\vec{a} + \vec{b} + \vec{c} ) = \vec{a} \times \vec{0} \\ \\ = > (\vec{a} \times \vec{a} ) + (\vec{a} \times \vec{b} ) + (\vec{a} \times \vec{c} ) = \vec{0}$

( °.° Product of any vector and zero vector is zero )

$= > \vec{0} + (\vec{a} \times \vec{b} ) + (\vec{a} \times \vec{c} ) = \vec{0}$

(°.° Cross product of same vectors is zero )

$= > (\vec{a} \times \vec{b}) - ( \vec{c} \times \vec{a}) = \vec{0} \\ \\ = > ( \vec{a} \times \vec{b} ) = (\vec{c} \times \vec{a} )$

Therefore,

we get,

$= > a.b \sin(\pi - C) = c.a \sin(\pi - B) \\ \\ = > b \sin(C) = c \sin(B) \\ \\ = > \frac{ \sin(C) }{c} = \frac{ \sin(B) }{b} \: \: \: \: \: .........(i)$

Similarly,

we can also prove that,

$= > \frac{ \sin(A) }{a} = \frac{ \sin(B) }{b} \: \: \: \: \: \: \: ............(ii)$

Therefore,

from eqn (i) and eqn (ii),

we conclude that,

$\large\bold{\frac{a}{ sin A } = \frac{b}{ sin B } = \frac{c}{ sin C }}$

Hence,

Proved

Answer 2

$\huge{\sf{\underline{\underline{\red{Solution:}}}}}$

To Prove:

$\bf{\implies \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}}$

So, as we know according to Triangle law of vector.

$\bf{\implies \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0}$

$\bf{\implies \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}}$

Now, cross product by $\bf{\overrightarrow{c}}$ both sides.

$\bf{\implies (\overrightarrow{a}+\overrightarrow{b}) \times \overrightarrow{c}=- \overrightarrow{c}\times \overrightarrow{c}}$

$\bf{\implies (\overrightarrow{a} + \overrightarrow{b})\times \overrightarrow{c}=0}$

$\bf{\implies \overrightarrow{a}\times \overrightarrow{c}+ \overrightarrow{b}\times \overrightarrow{c}=0}$

$\bf{\implies \overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{b}\times \overrightarrow{c}}$

$\bf{\implies |\overrightarrow{a}||\overrightarrow{c}|\sin (180^{\circ}-B)=|\overrightarrow{b}||\overrightarrow{c}|\sin (180^{\circ}-A)}$

$\bf{\implies a\sin B = b\sin A\;\;\;\;.............(1)}$

$\bf{\implies \dfrac{a}{\sin A}=\dfrac{b}{\sin B}}$

Similarly,

$\bf{\implies \overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}}$

$\bf{\implies (\overrightarrow{a}+\overrightarrow{c})\times \overrightarrow{b}=0}$

$\bf{\implies |\overrightarrow{a}||\overrightarrow{b}|\sin(180^{\circ}-c)=|\overrightarrow{b}||\overrightarrow{c}|\sin (180^{\circ}-A)}$

$\bf{\implies a\sin C=c\sin A}$

$\bf{\implies \dfrac{a}{\sin A}=\dfrac{c}{\sin C}\;\;\;\;\;.........(2)}$

From Eq (1) and Eq (2), we get

$\bf{\implies \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}}$

Hence Proved!!!!