Given sec θ = \( \dfrac{13}{12}\) Calculate all other trigonometric ratios

Asked by Abhisek | 1 year ago |  110

1 Answer

Solution :-

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right angled triangle ABC, right angled at B

sec θ =\( \dfrac{13}{12}\) 

\( \dfrac{Hypotenuse}{Adjacent \;side}=\dfrac{AC}{AB} \)

Let AC be 13k and AB will be 12k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is 

equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC2=AB+ BC2

Substitute the value of AB and AC

(13k)2= (12k)2 + BC2

169k2= 144k2 + BC2

169k2= 144k2 + BC2

BC2 = 169k2 – 144k2

BC2= 25k2

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = Opposite Side/Hypotenuse = \(\dfrac{BC}{AC} =\dfrac{5}{13} \)

Cos θ = Adjacent Side/Hypotenuse = \( \dfrac{AB}{AC} =\dfrac{12}{13} \)

tan θ = Opposite Side/Adjacent Side = \( \dfrac{BC}{AB} =\dfrac{5}{12} \)

Cosec θ = Hypotenuse/Opposite Side = \( \dfrac{AC}{BC} =\dfrac{13}{5} \)

cot θ = Adjacent Side/Opposite Side = \( \dfrac{AB}{BC} =\dfrac{12}{5} \)

Answered by Pragya Singh | 1 year ago

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