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Find the values of $x$ and $y$, and the area of the triangle:

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{tri(ans12,h1,lent11,ang11,ang12,w1)}

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$x$ = [[0]]mm

\n

$y$ = [[1]]mm

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Area = [[2]]$mm^2$

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First you need to find the third angle. Remember that the three angles of any triangle add up to 180 degrees. Next, pair up opposite sides and angles to use the sine rule.

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Find the values of $X,Y$ and $z$, and the area of the triangle:

\n

{tri2(w2,h2,ans23,lent21,lent22,ang21)}

\n

$X$ = [[0]] $^{\\circ}$

\n

$Y$ = [[1]] $^{\\circ}$

\n

$z$ = [[2]]mm

\n

Area = [[3]]mm$^2$

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Consider the sine rule and area formula below. Note that these formulae work for ANY triangle - it does not have to be right-angled. 

\n

\n

\n

\n

Note that the sine rule can also be rearranged to give:

\n

$\\frac{\\sin A}{a} = \\frac{\\sin B}{b} =\\frac{\\sin C}{c}$

\n

\n

\n

Now solve the following questions to 2 decimal places:

\n

\n

Note: You may need to scroll down to see the diagrams.

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(a)

\n

The missing angle is $180 - \\var{ang11} - \\var{ang12} = \\var{180-ang11-ang12}$

\n

Then we can use the sine rule to find $x$ and $y$:

\n

\n

$\\frac{\\var{lent11}}{\\sin(\\var{180-ang11-ang12})} = \\frac{x}{\\sin(\\var{ang12})}$

\n

$x =\\sin(\\var{ang12}) \\times \\frac{\\var{lent11}}{\\sin(\\var{180-ang11-ang12})} = \\var{ans11}$

\n

\n

$\\frac{\\var{lent11}}{\\sin(\\var{180-ang11-ang12})} = \\frac{y}{\\sin(\\var{ang11})}$

\n

$y =\\sin(\\var{ang11}) \\times \\frac{\\var{lent11}}{\\sin(\\var{180-ang11-ang12})} = \\var{ans12}$

\n

\n

To find the area we use the formula $\\text{Area} = \\frac{1}{2}ab\\sin C$

\n

\n

$\\text{Area} = \\frac{1}{2} \\times \\var{ans11} \\times \\var{lent11} \\times \\sin(\\var{ang11}^{\\circ}) = \\var{ans13}$ mm$^2$

\n

\n

\n

(b) 

\n

First use the rearranged sine rule to find $X$:

\n

$\\frac{\\sin(\\var{ang21})}{\\var{lent21}} = \\frac{\\sin(X)}{\\var{lent22}}$

\n

$\\sin(X) =\\var{lent22} \\times \\frac{\\sin(\\var{ang21})}{\\var{lent21}}$

\n

$X =\\sin ^{-1} (\\var{lent22} \\times \\frac{\\sin(\\var{ang21})}{\\var{lent21}}) = \\var{ans21}^{\\circ}$

\n

\n

Then we can find $Y$ by considering angles in a triangle:

\n

$Y = 180-\\var{ans21}-\\var{ang21} = \\var{ans22}^{\\circ}$

\n

\n

Finally use the sine rule again to find $z$:

\n

$\\frac{\\var{lent21}}{\\sin(\\var{ang21})} = \\frac{z}{\\sin(\\var{ans22})}$

\n

$z = \\frac{\\var{lent21}}{\\sin(\\var{ang21})} \\times \\sin(\\var{ans22}) = \\var{ans23}$ mm

\n

\n

To find the area we use the formula $\\text{Area} = \\frac{1}{2}ab\\sin C$:

\n

$\\text{Area} =\\frac{1}{2} \\times \\var{ans23} \\times \\var{lent22} \\times \\sin(\\var{ang21}) = \\var{ans24}$ mm$^2$

\n

\n

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Solve for x and y on a given triangle and calculate the area

\n

rebelmaths

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