Unit Circle: Trig Ratios
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Question 1 of 3
1. Question
Find `tan theta` given the following`sin theta=-21/29``cos theta``>``0`Write fractions in the format “a/b”- `tan theta=` (-21/20)
Hint
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Trigonometric Functions
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$
Pythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` are the two sides, and `c` is the hypotenuseFirst, draw a triangle that satisfies `sin theta=21/29` (the sign is disregarded in this process)
Using the triangle above, use the Pythagoras’ Theorem to solve for `x``a=21``b=x``c=29``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem `21^2``+``x^2` `=` `29^2` Substitute values `441+x^2` `=` `841` `441+x^2` `-411` `=` `841` `-411` Subtract `441` from both sides `x^2` `=` `400` `sqrt(x^2)` `=` `sqrt400` Get the square root of both sides `x` `=` `+-20` Recall the right triangle to find the value of `tan theta`
`tan theta` `=` $$\frac{\text{opposite}}{\text{adjacent}}$$ `=` `21/20` Remember that `tan theta=(sin theta)/(cos theta)``sin theta=-21/29``cos theta``>``0`Since `sin theta` is negative and `cos theta` is positive, `tan theta` is negativeTherefore, `tan theta=-21/20``tan theta=-21/20` -
Question 2 of 3
2. Question
Find `cos theta` given the following`sin theta=-3/5``(3pi)/2``<``theta``<``2pi`Write fractions in the format “a/b”- `cos theta=` (4/5)
Hint
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Incorrect
Trigonometric Functions
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$Pythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` are the two sides, and `c` is the hypotenuseFirst, draw a triangle that satisfies `sin theta=3/5` (the sign is disregarded in this process)
Using the triangle above, use the Pythagoras’ Theorem to solve for `x``a=3``b=x``c=5``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem `3^2``+``x^2` `=` `5^2` Substitute values `9+x^2` `=` `25` `9+x^2` `-9` `=` `25` `-9` Subtract `9` from both sides `x^2` `=` `16` `sqrt(x^2)` `=` `sqrt16` Get the square root of both sides `x` `=` `+-4` Recall the right triangle to find the value of `cos theta`
`cos theta` `=` $$\frac{\text{adjacent}}{\text{hypotenuse}}$$ `=` `4/5` Identify the appropriate quadrants in the unit circle to find the sign of `cos theta``(3pi)/2``<``theta``<``2pi`
In the fourth quadrant, `cos theta` is positiveTherefore, `cos theta=4/5``cos theta=4/5` -
Question 3 of 3
3. Question
Find `sin theta` given the following`cos theta=-5/13``pi``<``theta``<``(3pi)/2`Write fractions in the format “a/b”- `sin theta=` (-12/13)
Hint
Help VideoCorrect
Nice Job!
Incorrect
Trigonometric Functions
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$Pythagoras’ Theorem Formula
`a^2``+``b^2``=``c^2``a` and `b` are the two sides, and `c` is the hypotenuseFirst, draw a triangle that satisfies `cos theta=5/13` (the sign is disregarded in this process)
Using the triangle above, use the Pythagoras’ Theorem to solve for `x``a=y``b=5``c=13``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem `y^2``+``5^2` `=` `13^2` Substitute values `y^2+25` `=` `169` `y^2+25` `-25` `=` `169` `-25` Subtract `25` from both sides `y^2` `=` `144` `sqrt(y^2)` `=` `sqrt144` Get the square root of both sides `y` `=` `+-12` Recall the right triangle to find the value of `sin theta`
`sin theta` `=` $$\frac{\text{opposite}}{\text{hypotenuse}}$$ `=` `12/13` Identify the appropriate quadrants in the unit circle to find the sign of `cos theta``pi``<``theta``<``(3pi)/2`
In the third quadrant, `sin theta` is negativeTherefore, `sin theta=-12/13``sin theta=-12/13`