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Student is asked to find the distance from a given point, A, to a house, given the distance between A and another point B, and the angles at A and B. Requires use of the sine rule. Distance and angles are randomised.

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Using the fact that the angle sum of a triangle is \\(180^\\circ\\) we can find the angle at \\(H\\) to be \\begin{align} H&=180^\\circ -\\var{A}^\\circ -\\var{B}^\\circ\\\\&=\\var{180-A-B}^\\circ\\end{align}

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Now given that \\(AB=\\var{d}\\) m and using the sine rule we have \\begin{align}\\frac{b}{\\sin B}&=\\frac{h}{\\sin H}\\\\[5pt]\\frac{b}{\\sin \\var{B}}&=\\frac{\\var{d}}{\\sin \\var{180-A-B}}\\\\[5pt]b&=\\frac{\\var{d}\\sin \\var{B}}{\\sin \\var{180-A-B}}\\\\[5pt]&\\approx\\var{d*sin(pi*B/180)/sin(pi*(180-A-B)/180)}\\end{align}

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Hence the distance of the house from point \\(A\\) is \\(\\var{precround(d*sin(pi*B/180)/sin(pi*(180-A-B)/180),0)}\\) m to the nearest metre.

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distance between points A and B, random 500 to 900 in steps of 100

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Angle at A in degrees

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Angle at B in degrees

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Two points, \\(A\\) and \\(B\\), are \\(\\var{d}\\) metres apart with a house, \\(H\\), some distance away (see diagram). Given that the angle at \\(A\\) measures \\(\\var{A}^\\circ\\) and the angle at \\(B\\) measures \\(\\var{B}^\\circ\\) in the triangle \\(ABH\\), find the distance of the house from point \\(A\\) to the nearest metre. Write the distance in the box below and show full working on your written working.

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Distance = [[0]] m

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