McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Mid-Chapter Quiz
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Exercise 3 Page 829

Identify the hypotenuse as well as the adjacent and opposite sides of the angle. Use the Pythagorean Theorem to find the missing side length.

sin θ=sqrt(7)/4, cos θ=3/4, tan θ=sqrt(7)/3, csc θ=4sqrt(7)/7, sec θ=4/3, cot θ=3sqrt(7)/7

Practice makes perfect

In a right triangle the hypotenuse is the side that is opposite the right angle. If we take one of the acute angles as a reference, we can identify the opposite and adjacent sides to the angle.

In the given right triangle we have the hypotenuse and the adjacent side. However, we are missing the opposite side.
To find the opposite side length we will substitute b= 9 and c= 12 into the Pythagorean Theorem and solve for a.
a^2+b^2=c^2
a^2+ 9^2= 12^2
Solve for a
a^2 + 81 = 144
a^2 = 63
a=sqrt(63)
a = sqrt(9 * 7)
a = sqrt(9) * sqrt(7)
a = 3sqrt(7)
Note that when solving the equation we only kept the principal root. This is because a is the side length of a triangle and must be positive.

Let's find the values of the six trigonometric functions for angle θ. Remember to rationalize denominators if needed.

Function Substitute Simplify
sin θ=opp/hyp sin θ=3sqrt(7)/12 sin θ=sqrt(7)/4
cos θ=adj/hyp cos θ=9/12 cos θ=3/4
tan θ=opp/adj tan θ=3sqrt(7)/9 tan θ=sqrt(7)/3
csc θ=hyp/opp csc θ=12/3sqrt(7) csc θ=4sqrt(7)/7
sec θ=hyp/adj sec θ=12/9 sec θ=4/3
cot θ=adj/opp cot θ=9/3sqrt(7) cot θ=3sqrt(7)/7

Showing Our Work

Rationalizing Denominators
Rationalizing a denominator means eliminating the radical expression from the denominator. We had to rationalize the denominator of the expression 123sqrt(7). To do so, we expanded the fraction by multiplying the numerator and denominator by sqrt(7).
12/3sqrt(7)
4/sqrt(7)
4sqrt(7)/sqrt(7)* sqrt(7)
4sqrt(7)/(sqrt(7))^2
4sqrt(7)/7
We also followed this procedure to rationalize the denominator of 93sqrt(7).
9/3sqrt(7)
3/sqrt(7)
3sqrt(7)/sqrt(7)* sqrt(7)
3sqrt(7)/(sqrt(7))^2
3sqrt(7)/7