Can You Solve?? (1 Viewer)

brownie46 · Jul 9, 2017

Can anyone pls show working?

1729 · Jul 9, 2017

brownie46 said:
Can anyone pls show working?

$$\noindent \textbf{(a)} Since $\angle GKC = 45\degree$ then $\triangle GKC$ is an isosceles right-angled triangle, therefore, $CK = GC = h$. In $\triangle BCK$, $\angle BCK = 180\degree - 45\degree = 135 \degree$. Now apply the cosine rule, \\ \begin{align*} \quad BK^2 &= CK^2 + BC^2 - 2\left(CK\right)\left(BC\right) \cos{135\degree} \\ BK^2 &= h^2 + l^2 - 2hl\left(-\frac{\sqrt{2}}{2}\right) \\ BK^2 & = h^2 + l^2 + \sqrt{2}hl \end{align*}$

$$\noindent \textbf{(b)} In $\triangle BFK$, $\tan{30\degree} = \frac{h}{BK}$ and so $BK = \sqrt{3}h$. Substitute this into the result from part (a), \\ \begin{align*} \quad \left(\sqrt{3}h\right)^2 &= h^2 + l^2 + \sqrt{2}hl \\ 3h^2 &= h^2 + l^2 + \sqrt{2}hl \\ 2h^2 - l^2 &= \sqrt{2}hl \end{align*}$

$$\noindent \textbf{(c)} Rearranging the result from part (b), we have $2h^2 - \sqrt{2}hl - l^2 = 0$ which is a quadratic equation in $h$. Apply the quadratic formula, taking only the positive result, \\ \begin{align*} \quad h &= \frac{\sqrt{2}l + \sqrt{2l^2 - 4(2)(-l^2)}}{2(2)} \\ h &= \frac{\sqrt{2}l + \sqrt{10l^2}}{4} \\ h &= \frac{l\left(\sqrt{2} + \sqrt{10}\right)}{4} \\ \frac{h}{l} &= \frac{\sqrt{2}+\sqrt{10}}{4} \end{align*}$

brownie46 · Jul 10, 2017

Oh I see thanks dude, really helped!!

Can You Solve?? (1 Viewer)

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