Question

In any triangle ABC, prove that the area ∆= b²+c²-a²/4cot a​

Answer 1

$\bigstar \underline{\underline{\mathfrak{ To\ prove:-}}}$

• For any triangle ABC,  the area ∆= b²+c²-a²/4cot a​.

$\bigstar \underline{\underline{\mathfrak{Proof:-}}}$

We need to know:

$\leadsto \sf cotA =\frac{cosA}{sinA}$

$\leadsto \sf cosA=\frac{b^2+c^2-a^2}{2bc}$

Where,

a, b, c are the magnitudes of the sides opposite to the vertices A, B, C.

$\leadsto \sf \triangle = \frac{1}{2} bcsinA$

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LHS = Δ =1/2 bcsinA

RHS =  b²+c²-a²/4cot a​

We have to show that LHS = RHS .

Now,

RHS :

$:\implies \sf \frac{b^2+c^2-a^2}{4cotA}$

$:\implies \sf \frac{b^2 +c^2 -a^2}{4(cosA/sinA)}$

$:\implies \sf \frac{b^2+c^2-a^2 \times (sinA)}{cosA}$

$:\implies \sf \frac{\cancel{b^2+c^2-a^2}\times sinA}{4\times (\frac{\cancel{b^2+c^2-a^2}}{2bc})}$

$:\implies \sf \frac{2bc \times sinA}{4} \\\\:\implies \sf \frac{1}{2} bcsinA$

        = Δ

∴LHS =RHS

Hence proved!

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✪ Know More:-

$\mapsto \sf cosB=\frac{a^2+c^2-b^2}{2ac}$

$\mapsto \sf cosC=\frac{a^2+b^2-c^2}{2ab}$

$\mapsto \sf a^2=b^2+c^2 -2bccosA$

$\mapsto \sf b^2 =a^2 +c^2 -2ac\ cosB$

$\mapsto \sf c^2=b^2 +a^2 -2ab\ cosC$

$\mapsto \sf \frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC} \ \ (or)$

$\mapsto \sf \frac{sinA}{a} =\frac{sinB}{b} =\frac{sinC}{c}$

$\mapsto \sf cotA=\frac{cosA}{sinA} = \frac{1}{tanA}$

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Answer 2

$\bigstar \underline{\underline{\mathfrak{ To\ prove:-}}}$

For any triangle ABC,  the area ∆= b²+c²-a²/4cot a.

$\bigstar \underline{\underline{\mathfrak{Proof:-}}}$

We need to know:

$\leadsto \sf cotA =\frac{cosA}{sinA}$

$\leadsto \sf cosA=\frac{b^2+c^2-a^2}{2bc}$

Where,

a, b, c are the magnitudes of the sides opposite to the vertices A, B, C.

$\leadsto \sf \triangle = \frac{1}{2} bcsinA$

-------------------------

LHS = Δ =1/2 bcsinA

RHS =  b²+c²-a²/4cot a

We have to show that LHS = RHS .

Now,

RHS :

$:\implies \sf \frac{b^2+c^2-a^2}{4cotA}$

$:\implies \sf \frac{b^2 +c^2 -a^2}{4(cosA/sinA)}$

$:\implies \sf \frac{b^2+c^2-a^2 \times (sinA)}{cosA}$

$:\implies \sf \frac{\cancel{b^2+c^2-a^2}\times sinA}{4\times (\frac{\cancel{b^2+c^2-a^2}}{2bc})}$

$:\implies \sf \frac{2bc \times sinA}{4} \\\\:\implies \sf \frac{1}{2} bcsinA$

        = Δ

∴LHS =RHS

Hence proved!

------------------------

✪ Know More:-

$\mapsto \sf cosB=\frac{a^2+c^2-b^2}{2ac}$

$\mapsto \sf cosC=\frac{a^2+b^2-c^2}{2ab}$

$\mapsto \sf a^2=b^2+c^2 -2bccosA$

$\mapsto \sf b^2 =a^2 +c^2 -2ac\ cosB$

$\mapsto \sf c^2=b^2 +a^2 -2ab\ cosC$

$\mapsto \sf \frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC} \ \ (or)$

$\mapsto \sf \frac{sinA}{a} =\frac{sinB}{b} =\frac{sinC}{c}$

$\mapsto \sf cotA=\frac{cosA}{sinA} = \frac{1}{tanA}$

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