britbrat4290
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 7, b = 7, c = 5
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britbrat4290
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B. A = 69°, B = 69°, C = 42°??
mebs
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sounds good.
britbrat4290
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wrong
mebs
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Try Using cosine law
mebs
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\[7^{2} = 7^{2} + 5^{2} - 2(7)(5)cosA\]
britbrat4290
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my answers need to be in degrees
mebs
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you get \[cosA = \frac{ -25 }{ -70 }\] than do arc cosine. you get angle 69.07
britbrat4290
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i need degrees for a b n c ??
mebs
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you get angles 41.8 and 69.07. again
britbrat4290
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A. A = 70°, B = 70°, C = 40°
B. A = 69°, B = 69°, C = 42°
C. A = 42°, B = 69°, C = 69°
D. A = 69°, B = 42°, C = 69°
mebs
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|dw:1375984676609:dw| ?
mebs
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Try D.
mebs
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have to go.
britbrat4290
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thanks
britbrat4290
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deff not 69 deg
genius12
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Note that the triangle is isosceles:|dw:1375984026354:dw|
britbrat4290
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ok ? still confused
genius12
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Now use the cosine law:\[\bf a^2=b^2+c^2-2bc \cos(A) \implies 5^2=7^2+7^2-2(7)(7)\cos(A)\]Note that 'A' is the angle opposite side BC. Now re-arrange and solve for cos(A):\[\bf \implies \cos(A)=\frac{-73}{-98}=0.7445 \implies \cos^{-1}(0.7445)=A=41.884 \ degrees\]
genius12
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Now because the triangle is isosceles, the remaining two angles B and C are equal so we can find each of them by subtracting A from 180 and dividing the result by 2:\[\bf \angle B=\angle C=\frac{ 180 -41.884}{ 2 }=69.058 degrees\]Rounding off our results to the nearest degree we get:\[\bf \angle A =42 \ degrees\]\[\bf \angle B=\angle C=69 \ degrees\]
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genius12
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@britbrat4290 So your answer is correct =]