The base of a triangle is 80 and one of the b | vedclass

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(D) Let the sides of the triangle be $a, b, c$ where $a = 80$ and the angle $B = 60^{\circ}$.Given $b + c = 90$.Using the Law of Cosines: $b^2 = a^2 +

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$17$

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(D) Let the sides of the triangle be $a, b, c$ where $a = 80$ and the angle $B = 60^{\circ}$.Given $b + c = 90$.Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos(B)$.Substituting the values: $b^2 = 80^2 + c^2 - 2(80)(c) \cos(60^{\circ})$.Since $\cos(60^{\circ}) = 0.5$,we have $b^2 = 6400 + c^2 - 80c$.Substitute $b = 90 - c$ into the equation: $(90 - c)^2 = 6400 + c^2 - 80c$.$8100 - 180c + c^2 = 6400 + c^2 - 80c$.$1700 = 100c$.$c = 17$.Then $b = 90 - 17 = 73$.The sides are $80, 73, 17$. The shortest side is $17$.

The base of a triangle is $80$ and one of the base angles is $60^{\circ}$. If the sum of the lengths of the other two sides is $90$,then the shortest side is of length

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