Maths (1 Viewer)

highshill · Jul 9, 2017

Rectangle ABCD has a line BE drawn so that angle AEB equal 90 and DE equal 1 cm The width of the rectangle is 5 cm
A) find angle BEC
B) find the length of the rectangle

1729 · Jul 9, 2017

highshill said:
Rectangle ABCD has a line BE drawn so that angle AEB equal 90 and DE equal 1 cm The width of the rectangle is 5 cm
A) find angle BEC
B) find the length of the rectangle

$$\noindent \textbf{(a)} Produce $BA$ to $P$. (if the rectangle is named beginning from the top-left corner and going clockwise) \\ \begin{align*} \angle AEC &= \angle EAP \text{ (alternate angles)} \\ \text{But }\angle AEC &= 90 + \angle BEC \\ \text{and } \angle EAP &= 90 + \angle EAD \\ \therefore \angle EAD &= \angle BEC \\ \text{Now } \tan{\angle EAD} &= \frac{1}{5} \\ \angle EAD &= \tan^{-1}{\left(\frac{1}{5}\right)} \approx 11.3 \degree = \angle BEC \end{align*}$

$$\noindent \begin{align*} \textbf{(b) } \tan{\angle BEC} &= \frac{5}{EC} = \frac{1}{5} \\ EC &= 25 \\ \text{so length } l &= 25 + 1 = 26 \text{ cm} \end{align*}$

1729 · Jul 9, 2017

In (a) the construction of point P is not entirely necessary as long as you can prove angle EAD = angle BEC (which can be done multiple ways, such as alternate angles, angle sum of a triangle, angles on a straight line, etc)

highshill · Jul 9, 2017

Thanks 1729

Maths (1 Viewer)

highshill

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