Question

Given sec A =13/12, calculate all the other trigonometric ratios.​

Answer 1

And further you can calculate!!

Answer 2

Given that,

$\qquad \rm { • sec \: A = \dfrac{13}{12} }$

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We have to find,

$\qquad \rm { • All \: trigonometric \: ratios }$

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Solution,

We know that ,

$\qquad \rm { • sec \: A = \dfrac{Hypotenuse}{Base} }$

So,

$\qquad \rm { • Hypotenuse ( AC )= 13}$

$\qquad \rm { • Base ( AB )= 12}$

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Applying pythagoras property-

$\rm\red { {H}^{2} = {B}^{2} +{P}^{2} }$

$\rm { {P}^{2} = {H}^{2} - {B}^{2} }$

$\rm { \implies {P}^{2} = {H}^{2} - {B}^{2} }$

$\rm { \implies {P}^{2} = {13}^{2} - {12}^{2} }$

$\rm { \implies {P}^{2} = 169 - 144 }$

$\rm { \implies {P}^{2} = 25 }$

$\rm { \implies P = \sqrt{25}= 5 }$

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$\qquad \rm { • Hypotenuse (AC) = 13}$

$\qquad \rm { • Base ( AB ) = 12}$

$\qquad \rm { • Perpendicular (BC) = 5}$

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$\qquad \rm { • sin \:A = \dfrac{P}{H} = \dfrac{5}{13} }$

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$\qquad \rm { • cos \:A = \dfrac{B}{H} = \dfrac{12}{13} }$

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$\qquad \rm { • tan \:A = \dfrac{P}{B} = \dfrac{5}{12} }$

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$\qquad \rm { • cosec \:A = \dfrac{H}{P} = \dfrac{13}{5} }$

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$\qquad \rm { • sec \:A = \dfrac{H}{B} = \dfrac{13}{12} }$

$\:$

$\qquad \rm { • cot \:A = \dfrac{B}{P }= \dfrac{12}{5} }$