McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Trigonometric Functions in Right Triangles
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Exercise 3 Page 795

Draw a right triangle where cos A= 47. Use the trigonometric ratio to label two side lengths and the Pythagorean Theorem to find the missing side.

sin A=sqrt(33)/7, tan A=sqrt(33)/4, csc A=7sqrt(33)/33, sec A=7/4, cot A=4sqrt(33)/33

Practice makes perfect

In a right triangle, ∠ A is an acute angle. Cosine is defined as the ratio of the adjecent side to the hypotenuse of an angle. We are given that cos A= 47. cos = adj/hyp ⇒ cos A = 4 7 Now, let's draw a right triangle satisfying these conditions.

Let's find the opposite side by substituting b= 4 and c= 7 into the Pythagorean Theorem.

a^2+b^2=c^2
a^2 + 4^2 = 7^2
â–¼
Solve for a
a^2 + 16 = 49
a^2 = 33
a=sqrt(33)

Note that when solving the equation we only kept the principal root. This is because a represents a side length, and therefore, must be positive.

Now that we know the length of all three sides of the triangle, we can find the values of the five remaining trigonometric functions. Remember to rationalize denominators if needed.

Function Substitute Simplify
sin A=opp/hyp sin A=sqrt(33)/7 sin A=sqrt(33)/7
tan A=opp/adj tan A=sqrt(33)/4 tan A=sqrt(33)/4
csc A=hyp/opp csc A=7/sqrt(33) csc A=7sqrt(33)/33
sec A=hyp/adj sec A=7/4 sec A=7/4
cot A=adj/opp cot A=4/sqrt(33) cot A=4sqrt(33)/33

Showing Our Work

Rationalizing Denominators
Rationalizing a denominator means eliminating the radical expression from the denominator. Let's rationalize the denominator of the expression 7sqrt(33). To do so, we will expand the fraction by multiplying the numerator and denominator by sqrt(33).

7/sqrt(33)
7sqrt(33)/sqrt(33)* sqrt(33)
7sqrt(33)/(sqrt(33))^2
7sqrt(33)/33

Let's now follow the same procedure to rationalize the denominator of 4sqrt(33).

4/sqrt(33)
4sqrt(33)/sqrt(33) * sqrt(33)
4sqrt(33)/(sqrt(33))^2
4sqrt(33)/33