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c = 2, we get γ = arccos ◦)
53−28 cos(50◦ ) √7−2 cos(50 β = 50◦
c=2 radians ≈ 15.01◦ . We sketch the triangle below. a=7 α ≈ 114.99◦ γ ≈ 15.01◦ b ≈ 5.92 Now suppose instead of using the Law of Cosines to determine α, we use the Law of Sines.
Once b is determined, we have the angle-side opposite pair (β, b). Along with a, which is
given, we ﬁnd ourselves in the dreaded Angle-Side-Side (ASS) case. The Law of Sines gives
us sin(α) = sin(β ) , or sin(α) = a sin(β ) . Plugging in a = 7, β = 50◦ and b = 53 − 28 cos (50◦ ),
we get sin(α) = √ 7 sin(50 ) ◦ . The usual calculations yield the possibilities α ≈ 65.01◦ or
53−28 cos(50 ) α ≈ 180◦ − 65.01◦ = 114.99◦ . Both of these values for α are consistent with the angle-side
pair (β, b) in that there is more than enough room for either of these choices of α to reside
in a triangle with β = 50◦ , and both of these choices of α are greater than β , which agrees
with the observation that a > b. How...
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