How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=9, b=40, c=41?

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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=9, b=40, c=41?

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Answer 1 · 2018-02-16T08:23:09

See below:

Explanation:

Given that this is a right triangle, let's put side a along the positive $x$ $x$ axis, side b along the $y$ $y$ axis, and side c is the hypotenuse.

Now the question is Which angle are we concerned with? Is it angle A? Angle B? Or even the right angle, angle C?

Since we aren't given which angle, I'll work all 3.

The 6 trig functions are $sin, cos, tan, csc = \frac{1}{sin}, sec = \frac{1}{cos}, cot = \frac{1}{tan}$ $sin, cos, tan, csc = 1/sin, sec = 1/cos, cot = 1/tan$ (maybe you've learned SOHCAHTOA, or $sin = \frac{Opp}{Hyp}, cos = \frac{Adj}{Hyp}, tan = \frac{Opp}{Adj}$ $Sin = "Opp"/"Hyp", Cos = "Adj"/"Hyp", Tan = "Opp"/"Adj"$ ).

Angle A

Angle A has as its opposite side a, which means it's two sides are 1. along the $y$ $y$ axis and 2. the hypotenuse:

$sin = \frac{9}{41}; csc = \frac{41}{9}$ $sin=9/41; csc = 41/9$

$cos = \frac{40}{41}; sec = \frac{41}{40}$ $cos = 40/41; sec = 41/40$

$tan = \frac{9}{40}; cot = \frac{40}{9}$ $tan = 9/40; cot = 40/9$

Angle B

Angle B has as its opposite side b, which means it's two sides are 1. along the $x$ $x$ axis and 2. the hypotenuse:

$sin = \frac{40}{41}; csc = \frac{40}{41}$ $sin=40/41; csc = 40/41$

$cos = \frac{9}{41}; sec = \frac{41}{9}$ $cos = 9/41; sec = 41/9$

$tan = \frac{40}{9}; cot = \frac{9}{40}$ $tan = 40/9; cot = 9/40$

Angle C

Angle C is the right angle:

$sin = 1; csc = 1$ $sin=1; csc = 1$

$cos = 0; sec = undefined$ $cos = 0; sec = "undefined"$

$tan = undefined; cot = 0$ $tan = "undefined"; cot = 0$

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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=9, b=40, c=41?

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