Question

if cos A= 9\13 find csc A

Answer 1

Answer:

13$\sqrt{22}$ / 44

Step-by-step explanation:

cos A = 9/13

here adjacent = 9 and hypotenuse = 13  opposite = ?

using pythagoras theorem to find opposite

a^2 + b^2 = c^2

9^2 + b^2 = 13^2

81 + b^2 = 169

b^2 = 169 - 81

b^2 = 88

b = $\sqrt{88}$

b =  $2\sqrt{22}$

therefore opposite =  $2\sqrt{22}$

cosec A = hypotenuse/opposite

= 13/$2\sqrt{22}$

rationalizing the denominator

=13/ $2\sqrt{22}$  *  $2\sqrt{22}$ /  $2\sqrt{22}$

=13 *$2\sqrt{22}$ /( $2\sqrt{22}$ )^2

=26 $\sqrt{22}$ / 4*22

=26 $\sqrt{22}$ / 88

=13$\sqrt{22}$ / 44

Answer 2

Answer:

$cosec A = \frac{13}{\sqrt{88}}$

   OR

$cosec A = \frac{13 \sqrt{22}}{44}$

Step-by-step explanation:

Formulas used:

$cos^2 A = 1 - sin^2 A\\\\cosec A = \frac{1}{sin A}$

Given :

  $cos A = \frac{9}{13}$

Find cosec A

   $sin ^2 A = 1 - cos^2 A$

              $= 1 - (\frac{9}{13})^2\\\\= 1 - \frac{81}{169}\\\\=\frac{169 - 81}{169}\\\\=\frac{88}{169}$

 $sin A = \sqrt{\frac{88}{169}} = \frac{\sqrt{88}}{13}$

Therefore,

         $cosec A = \frac{1}{sin A} = \frac{1}{ \frac{\sqrt{88}}{13}} = \frac{13}{ \sqrt{88}}$

OR In a simplified form :

$cosec A = \frac{13}{\sqrt{88}} \times \frac{\sqrt{88}}{\sqrt{88}} = \frac {13 \times \sqrt{4 \times 22}}{88} = \frac{13 \times 2 \sqrt{22}} {88} = \frac{13 \sqrt{22}}{44}$