How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=5, b=12, c=13?

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Answer 1 · 2017-07-19T20:26:48

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So let's start with the angle $ˆ A$ $hatA$ :

$sin ˆ A = \frac{a}{c} = \frac{5}{13}$ $sin hatA=a/c=5/13$

$cos ˆ A = \frac{b}{c} = \frac{12}{13}$ $cos hatA=b/c=12/13$

$tan ˆ A = \frac{a}{b} = \frac{5}{12}$ $tan hatA=a/b=5/12$

$cot ˆ A = \frac{b}{a} = \frac{12}{5}$ $cot hatA=b/a=12/5$

$sec ˆ A = \frac{1}{cos ˆ A} = \frac{13}{12}$ $sec hatA=1/cos hatA=13/12$

$csc ˆ A = \frac{1}{sin ˆ A} = \frac{13}{5}$ $csc hatA=1/sin hatA=13/5$

Now for the angle $ˆ B$ $hatB$ :

$sin ˆ B = \frac{B}{c} = \frac{12}{13}$ $sin hatB=B/c=12/13$

$cos ˆ B = \frac{a}{c} = \frac{5}{13}$ $cos hatB=a/c=5/13$

$tan ˆ B = \frac{b}{a} = \frac{12}{5}$ $tan hatB=b/a=12/5$

$cot ˆ B = \frac{a}{b} = \frac{5}{12}$ $cot hatB=a/b=5/12$

$sec ˆ B = \frac{1}{cos ˆ B} = \frac{13}{5}$ $sec hatB=1/cos hatB=13/5$

$csc ˆ B = \frac{1}{sin ˆ B} = \frac{13}{12}$ $csc hatB=1/sin hatB=13/12$