Question

Given: In

∆ABC ,∠B=90° . If

sec=0

13

12

, calculate

all other

trigonometrical ratios.0​

Answer 1

Correct question :-

In ∆ABC , ∠B = 90° & sec∅ = 13/12. Then find all other trigonometric ratios

Solution :-

All trigonometric ratios

sin∅ = Opposite/Hypotenuse

cos∅ = Adjacent/Hypotenuse

tan∅ = Opposite/Adjacent

cosec∅ = Hypotenuse/Opposite

sec∅ = Hypotenuse/Adjacent

cot∅ = Adjacent/Opposite

Firstly finding other side of right ∆

If sec∅ = 13/12 , Hypotenuse of ∆ will be 13 & adjacent side of ∅ will be 12

Now , finding other side of ∅ using Pythagoras theorem

Hypotenuse² = Base² + Height²

13² = 12² + H²

169 - 144 = H²

H = √25

Height = 5

Now , finding all trigonometric ratios.

• Sin∅ = 5/13

• Cos∅ = 12/13

• Tan∅ = 5/12

• Cosec∅ = 13/5

• Sec∅ = 13/12

• Cot∅ = 12/5

Answer 2

Given that,

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

__________

We have to find,

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

__________

Solution,

We know that ,

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

So,

$\qquad \rm { • Hypotenuse = 13}$

$\qquad \rm { • Base = 12}$

___________

Applying pythagoras property-

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

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

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

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

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

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

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

____________

$\qquad \rm { • Hypotenuse = 13}$

$\qquad \rm { • Base = 12}$

$\qquad \rm { • Perpendicular = 5}$

____________

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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

$\:$

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