Question

Find the value of the other trigonometric ratios given that (1) cos theta = 5/13

Answer 1

Cos theta = 5/13

Base / Hypotenuse = 5/13

Base : Hypotenuse = 5 : 13

Let the base = 5a

Hypotenuse = 13a

Perpendicular = sqrt [ (13a)^2 - (5a)^2]

= sqrt ( 169a^2 - 25a^2)

= sqrt ( 144a^2)

= 12a

Sin theta = P/H = 12a / 13a = 12 /13

Cos theta = 5 /13

Tan theta = P / B = 12a / 5a = 12 / 5

Cosec theta = 1/sin theta = 13/12

Sec theta = 1/ cos theta = 13 /5

Cot theta = 1/ tan theta = 5 /12

Answer 2

Cos theta = 5/13 = B/H




B = 5 and H = 13.



By Pythagoras theroem,


( H )² = ( B)² + (P)²

( P)² = (H)² - (B)²



( P )² = (13)² - (5)²



P² = 144

P = √144


P = 12.




Therefore,



Sin theta = P/H = 12/13


Cos theta = B/H = 5/13


Tan theta = P/B = 12/5


Cosec theta = H/P = 13/12


Sec theta = H/B = 13/5


And,

Cot theta = B/P = 5/12