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Two questions testing the application of the Sine Rule when given two sides and an angle. In this question, the triangle is always acute and one of the given side lengths is opposite the given angle.
"}, "statement": "Suppose that $\\Delta ABC$ is a triangle with all interior angles $< \\dfrac{\\pi}{2}$ (in other words, an acute triangle). Here all angles are expressed in radians. Suppose also that standard naming conventions are used as indicated in the picture below (not necessarily an accurate picture of $\\Delta ABC$).
\nGiven the following angle and two side lengths, use the Sine Rule to determine the other side length and two angles. Write down the side length as a whole number and the angles (in radians) as decimals to 3dp.
\n\n\n", "advice": "a) We use the Sine Rule to find $B$: $\\dfrac{a}{\\sin A}=\\dfrac{b}{\\sin B}$. Thus $\\sin B=\\dfrac{b \\sin A}{a}=\\dfrac{\\var{b0}* \\var{s0}}{\\var{a0}}=\\var{b0*s0/a0}$. To find $B$ we need to calculate $\\sin^{-1} (\\var{b0*s0/a0})$, calculating the angle between $0$ and $\\dfrac{\\pi}{2}$, so $B=\\var{bb01}$ (to 3 decimal places).
\nSince $A+B+C=\\pi$, we calculate $C=\\pi-A-B=\\var{CC11}$. To 3dp, this gives $\\var{CC21}$.
\nWe use the Sine Rule to find $c$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$. Thus $c=\\dfrac{a \\sin C}{\\sin A}=\\dfrac{\\var{a0}* \\var{u21}}{\\var{s0}}=\\var{a0*u21/s0}$. The closest integer is then $\\var{c0}$. Note that this solution uses the 3dp value of $C$; the answer using $\\var{CC11}$ would give a slightly different long decimal value of $c$, but the integer value would be the same.
\nb) We use the Sine Rule to find $A$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$. Thus $\\sin A=\\dfrac{a \\sin C}{c}=\\dfrac{\\var{a3}* \\var{u3}}{\\var{c3}}=\\var{a3*u3/c3}$. To find $A$ we need to calculate $\\sin^{-1} (\\var{a3*u3/c3})$, calculating the angle between $0$ and $\\dfrac{\\pi}{2}$, so $A=\\var{aa31}$ (to 3 decimal places).
\nSince $A+B+C=\\pi$, we calculate $B=\\pi-A-C=\\var{bb41}$. To 3dp, this gives $\\var{bb51}$.
\nWe use the Sine Rule to find $b$: $\\dfrac{b}{\\sin B}=\\dfrac{c}{\\sin C}$. Thus $b=\\dfrac{c \\sin B}{\\sin C}=\\dfrac{\\var{c3}* \\var{t51}}{\\var{u3}}=\\var{c3*t51/u3}$. The closest integer is then $\\var{b3}$. Note that this solution uses the 3dp value of $B$; the answer using $\\var{bb41}$ would give a slightly different long decimal value of $b$, but the integer value would be the same.
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\nAngle $B=$ [[0]]
\nAngle $C=$ [[1]]
\nSide length $c=$ [[2]]
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\nAngle $A=$ [[0]]
\nAngle $B=$ [[1]]
\nSide length $b=$ [[2]]
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