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$A=\\var{AA0}$, $B=\\var{BB0}$, $a=\\var{a0}$

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Side length $b=$ [[0]]

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Angle $C=$ [[1]]

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Side length $c=$ [[2]]

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Use the Sine Rule to find $b$: $\\dfrac{a}{\\sin A}=\\dfrac{b}{\\sin B}$. Remember that $A+B+C=\\pi$. Use the Sine Rule to find $c$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$.

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$B=\\var{BB3}$, $C=\\var{CC3}$, $c=\\var{c3}$

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Side length $b=$ [[0]]

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Angle $A=$ [[1]]

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Side length $a=$ [[2]]

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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 $\\Delta ABC$).

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Given the following two angles and a side length, determine the other two side lengths and the angle. Write down the side lengths as whole numbers and the angle (in radians) as a decimal to 3dp.

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\n \n \n ", "tags": ["checked2015", "SFY0001", "sine rule", "Sine Rule", "Solving triangles", "Triangle", "Two angles and a side"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "

I want acute triangles with side lengths $a,b,c$. I need $|a^2-b^2|<c^2<a^2+b^2$ along with corresponding conditions on $a,b$. In fact the conditions $a^2-b^2<c^2<a^2+b^2$ and $b^2-a^2<c^2<a^2+b^2$ imply also the corresponding conditions on $a,b$. Thus the design of the question involves choosing $a,b$ and then choosing $c$ to meet the required condition. The integer $c$ is chosen randomly between the ceiling of $\\sqrt{|a^2-b^2|}$ and the floor of $\\sqrt{a^2+b^2}$. The first is no greater than the second because $\\max\\{a,b\\}$ lies between them; if $a=b$, then $\\sqrt{a^2+b^2} > 1$. The range of values for $a$ and $b$ may be changed according to taste without invalidating the question, but questions arise about accuracy. My calculations suggest that values of $a,b,c$ between 5 and 100 are safe, but I have been more conservative than that.

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The second part tests the ability to apply the same principles as the first part but with a different orientation to the triangle: the first part seeks $b,C,c$ whereas the second seeks $b,A,a$.

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Two questions testing the application of the Sine Rule when given two angles and a side. In this question, the triangle is always acute.

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a) We use the Sine Rule to find $b$: $\\dfrac{a}{\\sin A}=\\dfrac{b}{\\sin B}$. Thus $b=\\dfrac{a \\sin B}{\\sin A}=\\dfrac{\\var{a0}* \\var{t0}}{\\var{s0}}=\\var{a0*t0/s0}$. The closest integer is then $\\var{b0}$.

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Since $A+B+C=\\pi$, we calculate $C=\\pi-A-B=\\var{CC1}$. To 3dp, this gives $\\var{CC2}$.

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We use the Sine Rule to find $c$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$. Thus $c=\\dfrac{a \\sin C}{\\sin A}=\\dfrac{\\var{a0}* \\var{u2}}{\\var{s0}}=\\var{a0*u2/s0}$. The closest integer is then $\\var{c0}$. Note that this solution uses the 3dp value of $C$; the answer using $\\var{CC1}$ would give a slightly different long decimal value of $c$, but the integer value would be the same.

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b) We use the Sine Rule to find $b$: $\\dfrac{b}{\\sin B}=\\dfrac{c}{\\sin C}$. Thus $b=\\dfrac{c \\sin B}{\\sin C}=\\dfrac{\\var{c3}* \\var{t3}}{\\var{u3}}=\\var{c3*t3/u3}$. The closest integer is then $\\var{b3}$.

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Since $A+B+C=\\pi$, we calculate $A=\\pi-B-C=\\var{AA4}$. To 3dp, this gives $\\var{AA5}$.

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We use the Sine Rule to find $a$: $\\dfrac{a}{\\sin A}=\\dfrac{c}{\\sin C}$. Thus $a=\\dfrac{c \\sin A}{\\sin C}=\\dfrac{\\var{c3}* \\var{s5}}{\\var{u3}}=\\var{c3*s5/u3}$. The closest integer is then $\\var{a3}$. Note that this solution uses the 3dp value of $A$; the answer using $\\var{AA4}$ would give a slightly different long decimal value of $a$, but the integer value would be the same.

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