Stitz-Zeager_College_Algebra_e-book

To solve this equation we need the arccosine function

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hich 4 2 also contains α = 30◦ . This means that γ must measure between 0◦ and 150◦ in order to fit inside the triangle with α. The only angle that satisfies this requirement and has sin(γ ) = 1 is γ = 90◦ . In other words, we have a right triangle. We find the measure of β to be β =◦ 180◦ − 30◦ − 90◦ = 60◦ and then determine b using the Law of Sines. We find √ sin(60 b = 2sin(30◦ )) = 2 3 ≈ 3.46 units. In this case, the side a is precisely long enough to form a unique right triangle. c=4 a=1 β = 60◦ c=4 α = 30◦ a=2 α = 30◦ b ≈ 3.46 Diagram for number 3 Triangle for number 4 5. Proceeding as we have in the previous two examples, we use the Law of Sines to find γ . In this ◦ ◦ case, we have sin(γ ) = sin(30 ) or sin(γ ) = 4 sin(30 ) = 2 . Since γ lies in a triangle with α = 30◦ , 4 3 3 3 we must have that 0◦ < γ < 150◦ . There are two angles γ that fall in this range and have sin(γ ) = 2 : γ = arcsin 2 radians ≈ 41.81◦ and γ = π − arcsin 2 radians ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online