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Question
In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.
sinB = `sqrt(3)/(2)`
Solution
sinB = `sqrt(3)/(2)`
sinB = `"Perpendicular"/"Hypotenuse" = sqrt(3)/(2)`
By Pythagoras theorem, we have
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
⇒ Base = `sqrt(("Hypotenuse")^2 - ("Perpendicular")^2`
⇒ Base
`sqrt((2)^2 - (sqrt(3))^2`
= `sqrt(4 - 3)`
= `sqrt(1)`
= 1
cosB = `"Base"/"Hypotenuse" = (1)/(2)`
tanB = `"Perpendicular"/"Base" = sqrt(3)`
secB = `(1)/"cosB"` = 2
cotB = `(1)/"tanB" = (1)/sqrt(3)`
cosecB= `(1)/"sinA" = (2)/sqrt(3)`.
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