Stitz-Zeager_College_Algebra_e-book

# 2 example 1045 1 use a half angle formula to nd the

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Unformatted text preview: ow that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦ . We now proceed to ﬁnd the lengths of the remaining two sides of the triangle. Let c denote the 7 length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30◦ ) = 7 , or c = cos(30◦ ) . c √ √ Since cos (30◦ ) = 23 , we have, after the usual fraction gymnastics, c = 143 3 . At this point, we have two ways to proceed to ﬁnd the length of the side opposite the 30◦ angle, which we’ll denote √ b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 143 3 , so we could use the Pythagorean Theorem to ﬁnd the missing side and solve (7)2 + b2 = √2 14 3 3 for b. b Alternatively, we could use Theorem 10.4, namely that sin (30◦ ) = c . Choosing the latter, we ﬁnd b = c sin (30◦ ) = √ 14 3 3 · 1 2 = √ 73 3. The triangle with all of its data is recorded below. 10.2 The Unit Circle: Cosine and Sine c= √ 14 3 3 629 60◦ b= √ 73 3 30◦ 7 We close this section by noting that we can easily extend the functions cosine and sine to real...
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