Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
3. Right Triangles and Trigonometric Ratios
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Exercise 31 Page 925

In a right triangle, the cotangent of an acute angle is defined as the ratio of the length of the adjacent side to the angle to the length of the opposite side to the angle.

Triangle:

Trigonometric Ratios: sin θ=4sqrt(41)/41, cos θ=5sqrt(41)/41, tan θ=4/5, csc θ=sqrt(41)/4, sec θ=sqrt(41)/5

Practice makes perfect

Given that cot θ= 54, 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.

Drawing the Triangle

In a right triangle the cotangent of an acute angle is defined as the ratio of the length of the adjacent side to the angle to the length of the opposite side to the angle. cot θ =5/4 ⇔ cot θ = adjacent/opposite

Therefore, we know that the length of the adjacent side to θ is 5 and that the length of the opposite side to θ is 4.

We can find the length of the hypotenuse by substituting a= 5 and b= 4 into the Pythagorean Theorem.
a^2+b^2=c^2
5^2+ 4^2=c^2
Solve for a
25+16=c^2
41=c^2
sqrt(41)=c
c= sqrt(41)
Note that when solving the equation we only considered the principal root. This is because c represents a side length and therefore must be a positive number. We can now draw the right triangle and label its three sides.

Finding Trigonometric Ratios

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
sin θ=opp/hyp sin θ=4/sqrt(41) sin θ=4sqrt(41)/41
cos θ=adj/hyp cos θ=5/sqrt(41) cos θ=5sqrt(41)/41
tan θ=opp/adj tan θ=4/5 -
csc θ=hyp/opp csc θ=sqrt(41)/4 -
sec θ=hyp/adj sec θ=sqrt(41)/5 -

Showing Our Work

Rationalizing Denominators
Rationalizing a denominator means eliminating any radical expression from the denominator. In the work above we needed to rationalize the denominators of two expressions, 4sqrt(41) and 5sqrt(41). Let's look at how this was done for 4sqrt(41) first.
4/sqrt(41)
4sqrt(41)/sqrt(41)*sqrt(41)
4sqrt(41)/41
Let's now follow the same procedure to rationalize the denominator of5sqrt(41).
5/sqrt(41)
5sqrt(41)/sqrt(41)*sqrt(41)
5sqrt(41)/41