Question
In a $\triangle\text{ABC},$ if a = 8, b = 10, c = 12 and $\text{C}=\lambda\text{A},$ find the value of $\lambda.$

Answer

Using cosine rule, we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\Rightarrow\cos\text{A}=\frac{10^2+12^2-8^2}{2\times10\times12}$
$\Rightarrow\cos\text{A}=\frac{100+144-64}{240}$
$\Rightarrow\cos\text{A}=\frac{180}{240}=\frac{3}{4}...(1)$
Now, using sine rule, we have
$\Rightarrow\frac{\text{a}}{\sin\text{A}}=\frac{\text{c}}{\sin\text{C}}$
$\Rightarrow\frac{8}{\sin\text{A}}=\frac{12}{\sin\lambda\text{A}}$
$\Rightarrow\sin\lambda\text{A}=\frac{3}{2}\sin\text{A}$
$\Rightarrow\sin\lambda\text{A}=2\times\frac{3}{4}\sin\text{A}$
$\Rightarrow\sin\lambda\text{A}=2\sin\text{A}\cos\text{A}$ [Using (1)]
$\Rightarrow\sin\lambda\text{A}=\sin2\text{A}$
$\Rightarrow\lambda=2$

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