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Question
In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.
sinA = `(12)/(13)`
Solution
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)`.
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