Question

3cotA=2 எ‌னி‌ல் (4sinA-3cosA)/(2sinA+3cosA) இ‌ன் ம‌தி‌ப்பை‌க் கா‌ண்க

Answer 1

Step-by-step explanation:

3cotA=2

cotA=2/3

hypotenuse= √9+√4

√13

Put it in question

Answer 2

‌விள‌க்க‌ம்:

3 cot A=2

$\cot A=\frac{2}{3}$

$\cot A=\frac{\text { adj side }}{\text { opp side }}=\frac{2}{3}$

$A C^{2}=A B^{2}+B C^{2}$

       $=2^{2}+3^{2}$

$A C=\sqrt{4+9}$

      $=\sqrt{13}$

$\sin A=\frac{\text {opp.side}}{\text {hypo}}=\frac{3}{\sqrt{13}}$

$\frac{4 \sin A-3 \cos A}{2 \sin A+3 \cos A}$

$\Rightarrow \frac{4\left(\frac{3}{\sqrt{13}}\right)-3\left(\frac{2}{\sqrt{13}}\right)}{2\left(\frac{3}{\sqrt{3}}\right)+3\left(\frac{2}{\sqrt{13}}\right)}$

$\Rightarrow \frac{\frac{12}{\sqrt{13}} - \frac{6}{\sqrt{13}}}{\frac{6}{\sqrt{13}}+\frac{6}{\sqrt{13}}}$

$\Rightarrow \frac{\frac{12-6}{\sqrt{13}}}{\frac{6+6}{\sqrt{13}}}$

$\Rightarrow \frac{6}{12}=\frac{1}{2}$

$\frac{4 \sin A-3 \cos A}{2 \sin A+3 \cos A}=\frac{1}{2}\right.$

ம‌தி‌ப்பு = $\frac{1}{2}$