Question

if sin A=12/13 and (2 sinA-k cos A) /(4sinA-9cosA) =3 find the value of k​

Answer 1

Answer:

$\sin( \alpha ) = \frac{12}{13} = \frac{p}{h}$

So,

$b = \sqrt{ {13}^{2} - {12}^{2} } = \sqrt{25} = 5$

i.e,

$\cos( \alpha ) = \frac{5}{13}$

now putting the value of Sin & Cos in question

$\frac{2 \sin\alpha - k \: cos \: \alpha }{4sin \: \alpha - 9 \: cos \: \alpha } = \frac{2 \times \frac{12}{13} - k \times \frac{5}{13} }{4 \times \frac{12}{13} - 9 \times \frac{5}{13} } = 3$

$\frac{24 - 5k}{48 - 45} = 3$

$24 - 5k = 3 \times 3 = 9$

$5k = 24 - 9 = 15$

$k = \frac{15}{5} = 3 \: \: \: ans.$