Stitz-Zeager_College_Algebra_e-book

# Since all of the equations in this example are found

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Unformatted text preview: lution. We sketch the problem below with the ﬁrst observation point labeled as P and the second as Q. In order to use the Law of Signs to ﬁnd the distance d from Q to the island, we ﬁrst need to ﬁnd the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45◦ are supplemental, so that γ = 180◦ − 45◦ = 135◦ . We can now ◦ ◦ ﬁnd β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦ . By the Law of Sines, we have sin(30 ) = sin(15 ) d 5 sin(30◦ which gives d = 5sin(15◦ )) ≈ 9.66 miles. Next, to ﬁnd the point on the coast closest to the island, which we’ve labeled as C , we need to ﬁnd the perpendicular distance from the island to the coast.10 9 10 Remember, we have already argued that a triangle exists in this case! Do you see why C must lie to the right of Q? 768 Applications of Trigonometry Let x denote the distance from the second observation point Q to the point C and let y denote the distance...
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