Question

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

sec B = `(15)/(12)`

Answer 1

sec B = `(15)/(12)`

sec B = `(1)/"cos B" = "Hypotenuse"/"Base" = (15)/(12)`

By Pythagoras theorem, we have
(Hypotenuse)² = (Perpendicular)² + (Base)²
⇒ Perpendicular = `sqrt(("Hypotenuse")^2 - ("Base")^2`
⇒ Perpendicular
= `sqrt((15)^2 - (12)^2`
= `sqrt(225 - 144)`
= `sqrt(81)`
= 9

sinB = `"Perpendicular"/"Hypotenuse" = (9)/(15)`

tanB = `"Perpendicular"/"Base" = (9)/(12)`

cotB = `(1)/"tan B" = (12)/(9)`

cosecB = `(1)/"sinB" = (15)/(9)`

cos B = `"Base"/"Hypotenuse" = (12)/(15)`.