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In a right triangle, the sine of an acute angle is defined as the ratio of the opposite side to the hypotenuse.
Triangle:
Trigonometric Ratios: cos θ=3sqrt(5)/7, tan θ=2sqrt(5)/15, csc θ=7/2, sec θ=7sqrt(5)/15, cot θ=3sqrt(5)/2
Given that sin θ= 27, we want to sketch a right triangle with θ as the measure of one acute angle. Then, we will find the other five trigonometric ratios of θ. Let's do these things one at a time.
In a right triangle, the sine of an acute angle is defined as the ratio of the opposite side to the hypotenuse.
sin θ =2/7 ⇔ sin θ = opposite/hypotenuse
We can find the missing leg length by substituting b= 2 and c= 7 into the Pythagorean Theorem.
Note that when solving the equation we only considered the principal root. This is because a represents a side length and therefore must be a positive number. We can now draw the right triangle and label its three sides.
Having the three sides of the right triangle allows us to find the five remaining trigonometric ratios. Remember to rationalize denominators, if needed.
| Function | Substitute | Simplify |
|---|---|---|
| cos θ=adj/hyp | cos θ=3sqrt(5)/7 | - |
| tan θ=opp/adj | tan θ=2/3sqrt(5) | tan θ=2sqrt(5)/15 |
| csc θ=hyp/opp | csc θ=7/2 | - |
| sec θ=hyp/adj | sec θ=7/3sqrt(5) | sec θ=7sqrt(5)/15 |
| cot θ=adj/opp | cot θ=3sqrt(5)/2 | - |
a/b=a * sqrt(5)/b * sqrt(5)
a* a=a^2
( sqrt(a) )^2 = a
Multiply
Let's now follow the same procedure to rationalize the denominator of 73sqrt(5).
a/b=a * sqrt(5)/b * sqrt(5)
a* a=a^2
( sqrt(a) )^2 = a
Multiply