How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=20, b=21, c=29?

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How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=20, b=21, c=29?

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Answer 1 · 2018-02-19T21:56:31

Use the identities.

Explanation:

The answers you come up with are going to depend on whether you're evaluating $∠ A$ $angleA$ or $∠ B$ $angleB$ . For the sake of this answer, we can look at $∠ A$ $angleA$ .

We are given the lengths of all sides. $c$ $c$ is the side across from $∠ C$ $angleC$ , $a$ $a$ across from $∠ A$ $angleA$ , and $b$ $b$ across from $∠ B$ $angleB$ .

Drawing a model at this point is helpful.

Image courtesy of me and Microsoft Paint.

We know the identities of the three normal trig functions:

SOH
$sin θ = \frac{o p p o s i t e}{h y p o t e n u s e}$ $sintheta = (opposite)/(hypoten\use)$

CAH
$cos θ = \frac{a d j a c e n t}{h y p o t e n u s e}$ $costheta = (adjacent)/(hypoten\use)$

TOA
$tan θ = \frac{o p p o s i t e}{a d j a c e n t}$ $tantheta = (opposite)/(adjacent)$

We can just substitute in our values:

$sin A = \frac{20}{29}$ $sinA = 20/29$
$cos A = \frac{21}{29}$ $cosA = 21/29$
$tan A = \frac{20}{21}$ $tanA = 20/21$

And flip them for the inverse functions.

$csc A = \frac{29}{20}$ $cscA = 29/20$
$sec A = \frac{29}{21}$ $secA = 29/21$
$cot A = \frac{21}{20}$ $cotA = 21/20$

We could easily do the same for $∠ B$ $angleB$ , flipping our values for $o p p o s i t e$ $opposite$ and $a d j a c e n t$ $adjacent$ .

$sin B = \frac{21}{29}$ $sinB = 21/29$
$cos B = \frac{20}{29}$ $cosB = 20/29$
$tan B = \frac{21}{20}$ $tanB = 21/20$
$csc B = \frac{29}{21}$ $cscB = 29/21$
$sec B = \frac{29}{20}$ $secB = 29/20$
$cot B = \frac{20}{21}$ $cotB = 20/21$

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How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=20, b=21, c=29?

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