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Question
In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.
sec B = `(15)/(12)`
Solution
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)`.
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