Question

In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.

sinA = `(12)/(13)`

Answer 1

sinA = `(12)/(13)`

sinA = `"Perpendicular"/"Hypotenuse" = (12)/(13)`

By Pythagoras theorem, we have
(Hypotenuse)² = (Perpendicular)² + (Base)²
⇒ Base = `sqrt(("Hypotenuse")^2 - ("Perpendicular")^2`
⇒ Base
= `sqrt((13)^2 - (12)^2`
= `sqrt(169 - 144)`
= `sqrt(25)`
= 5

cosA = `"Base"/"Hypotenuse" = (5)/(13)`

secA = `(1)/"cosA" = (13)/(5)`

cotA = `(1)/"tanA" = (5)/(12)`

cosecA = `(1)/"sinA" = (13)/(12)`.