How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=1, b=2, $c=sqrt5$ ?

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Answer 1 · 2016-12-12T20:04:46

Please see the explanation.

Given: $a = 1, b = 2, c = sqrt(5)$

$sin(A) = a/c = sqrt(5)/5 = cos(B)$

$csc(A) = c/a = sqrt(5) = sec(B)$

$cos(A) = b/c = (2sqrt(5))/5 = sin(B)$

$sec(A) = c/b = sqrt(5)/2 = csc(B)$

$tan(A) = a/b = 1/2 = cot(B)$

$cot(A) = b/a = 2 = tan(B)$

How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=1, b=2, c=sqrt5c=sqrt5?