Question

13
5. Given sec 0 =
12
calculate all other trigonometric ratios.​

Answer 1

Given that,

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

__________

We have to find,

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

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Solution,

We know that ,

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

So,

$\qquad \rm { • Hypotenuse = 13}$

$\qquad \rm { • Base = 12}$

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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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$\qquad \rm { • Hypotenuse = 13}$

$\qquad \rm { • Base = 12}$

$\qquad \rm { • Perpendicular = 5}$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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