Question

In triangle ABC, right angled at C, find cos A, tan A and cosec B, if sin A =24/25.​

Answer 1

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Answer 2

$\sin(a) = \dfrac{24}{25} \\ \\ \frac{p}{h} = \dfrac{24}{25} \\ \\ let \: p \: be \: 24k \: and \: h \: be \: 25k \\ \\ since \\ \\ the \: triangle \: is \: a \: right \: angled \: triangle \\ \\ applying \: pythagoras \: theorem \\ \\ {p}^{2} + {b}^{2} = {h}^{2} \\ \\ {b}^{2} = {h}^{2} - {p}^{2} \\ \\ {b}^{2} = {(25k)}^{2} - {(24k)}^{2} \\ \\ {b}^{2} = 625 {k}^{2} - 576 {k}^{2} \\ \\ b = \sqrt{49 {k}^{2} } = 7k$

Now,

$\cos(a) = \dfrac{b}{h} \\ \\ \cos(a) = \frac{ac}{ab} = \frac{7k}{25k} = \frac{7}{25}$


Now,

$\tan(a) = \frac{p}{b} \\ \\ \tan(a) = \frac{bc}{ac} = \frac{24k}{7k} = \frac{24}{7}$


Now,

$\cosec(b) = \frac{ab}{ac} = \frac{25k}{7k} = \frac{25}{7}$


Thanks!