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"Given that $a=4$, $b=5$, and the area of $\\triangle ABC$ is $5 \\sqrt {3}$, I can use the formula for the area of a triangle in terms of its side lengths.\n\nThe area of a triangle can be expressed as $$A = \\frac{1}{2}ab\\sin C,$$ where $A$ is the area, $a$ and $b$ are the lengths of two sides, and $C$ is the an... | [
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"$$b^2 = 16 + 31 - 2\\sqrt {31} \\left( \\frac{26}{40}",
"However, this equation is not true, so I must have made a mistake",
"However, this equation is not true, so I must have made a ... | [
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] | In acute triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. If $a=4$, $b=5$, and the area of $\triangle ABC$ is $5 \sqrt {3}$, find $|AB|$. | I need to find the length of side $AB$, which is $b$ in the given triangle.
| 0 | 0 | 0 | \\sqrt{21} |
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"Given that the area of $\\triangle ABC$ is $5\\sqrt{3}$ and $a=4$, $b=5$, I can substitute these values into the formula for the area:\n$$5\\sqrt{3} = \\frac{1}{2}(4)(5)\\sin C.$$\n\nSimplifying the equation:\n$$5\\sqrt{3} = 10\\sin C.$$\n\nDividing both sides by 10:\n$$\\frac{5\\sqrt{3}}{10} = \\sin C.$$\n\nSimpl... | [
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"$$\\frac{\\frac{625 \\cdot 3^2}{16\\sin^4 C} - \\frac{18 \\cdot 25 \\cdot 3}{\\sin^2 C} + 81}{64 \\cdot \\frac{25 \\cdot 3}{\\sin^2 C}} + \\frac{12}{b^2}",
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] | In acute triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. If $a=4$, $b=5$, and the area of $\triangle ABC$ is $5 \sqrt {3}$, find $|AB|$. | I need to find the length of side $AB$, which is $b$ in the given triangle.
I know that the area of a triangle can be expressed as $$\frac{1}{2}ab\sin C$$, where $a$ and $b$ are the lengths of two sides of the triangle and $C$ is the angle between them.
| 0 | 0 | 1 | \\sqrt{21} |
[
"However, I don't know the angle $C$, but I can use the fact that the area of a triangle can also be expressed as $$\\frac{1}{2}bc\\sin A$$, where $A$ is the angle opposite to side $a$.\n\nEquating these two expressions for the area, I get $$\\frac{1}{2}ab\\sin C = \\frac{1}{2}bc\\sin A.$$\n\nSince $a = 4$ and $b =... | [
"Now, I can use the Law of Cosines to write $$5^2 = 4^2 + \\left( \\frac{10 \\sqrt {3}}{\\sin C} \\right )^2 - 2(4)\\left( \\frac{10 \\sqrt {3}}{\\sin C} \\right )\\cos B = 16 + \\frac{300}{\\sin^2 C} - \\frac{80 \\sqrt {3}}{\\sin C}\\cos B.$$",
"Substituting, I get $$5 \\sqrt {3} = \\frac{10c}{\\frac{5c}{\\sin A... | [
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] | In acute triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. If $a=4$, $b=5$, and the area of $\triangle ABC$ is $5 \sqrt {3}$, find $|AB|$. | I need to find the length of side $AB$, which is $b$ in the given triangle.
I know that the area of a triangle can be expressed as $$\frac{1}{2}ab\sin C$$, where $a$ and $b$ are the lengths of two sides of the triangle and $C$ is the angle between them.
In this case, the area of the triangle is given as $5 \sqrt {3}$... | 0 | 0 | 2 | \\sqrt{21} |
[
"Simplifying the equation: $$5 \\sqrt {3} = 10 \\sin C.$$\n\nDividing both sides by 10: $$\\frac{5 \\sqrt {3}}{10} = \\sin C.$$\n\nThis implies that $$\\sin C = \\frac{\\sqrt {3}}{2}.$$\n\nSince $\\sin C = \\frac{\\sqrt {3}}{2}$, I know that $C = 60^\\circ$.\n\nNow, I can use the Law of Cosines to relate the length... | [
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"Since $\\frac{8}{\\sqrt {3}} = \\frac{8}{\\sqrt {3}} \\cdot \\frac{\\sqrt {3}}{\\sqrt {3}} = \\frac{8\\sqrt {3}}{3}$, and $\\frac{15}{\\sqrt {3}} = \\frac{15}{\\sqrt {3",
"Since $\\frac{8}{\\sqrt {3}} = \\frac{8}{\\sqrt {3}} \\cdot \\frac{\\sqrt {3}}{\\sqrt {3}} = \\frac{8\\sqrt {3}}{3}$, and $\\frac{15}{... | [
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] | In acute triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. If $a=4$, $b=5$, and the area of $\triangle ABC$ is $5 \sqrt {3}$, find $|AB|$. | I need to find the length of side $AB$, which is $b$ in the given triangle.
I know that the area of a triangle can be expressed as $$\frac{1}{2}ab\sin C$$, where $a$ and $b$ are the lengths of two sides of the triangle and $C$ is the angle between them.
In this case, the area of the triangle is given as $5 \sqrt {3}$... | 0 | 0 | 3 | \\sqrt{21} |
[
"Dividing both sides by 10, I find that $$\\sin C = \\frac{\\sqrt {3}}{2}.$$\n\nSince $C$ is an acute angle, I can see that $C = 60^\\circ$ (or $\\frac{\\pi}{3}$ radians).\n\nNow, I can use the Law of Sines to find the length of side $b$: $$\\frac{b}{\\sin B} = \\frac{a}{\\sin A}.$$\n\nSince $A = 90^\\circ - C = 30... | [
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"Taking the square... | [
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] | In acute triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. If $a=4$, $b=5$, and the area of $\triangle ABC$ is $5 \sqrt {3}$, find $|AB|$. | I need to find the length of side $AB$, which is $b$ in the given triangle.
I know that the area of a triangle can be expressed as $$\frac{1}{2}ab\sin C$$, where $a$ and $b$ are the lengths of two sides of the triangle and $C$ is the angle between them.
In this case, the area of the triangle is given as $5 \sqrt {3}$... | 0 | 0 | 4 | \\sqrt{21} |
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