## Challenging Trigonometry Question (ACS(I) Sec 3)

Solution:

(a) $\tan (\alpha + \beta )=\frac{10}{x}$

Using the formula,

$\displaystyle \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}=\frac{10}{x}$

$\displaystyle \frac{\frac{5}{x}+\tan \beta}{1-(\frac{5}{x})\tan \beta}=\frac{10}{x}$

Cross-multiply,

$5+x\tan\beta =10-\frac{50}{x}\tan \beta$

$(x+\frac{50}{x})\tan\beta =5$

$\displaystyle \tan \beta =\frac{5}{x+\frac{50}{x}}=\frac{5x}{x^2+50}$

(b) The trick here is to break up $2\alpha +\beta$ into $\alpha + (\alpha +\beta )$

$\displaystyle \begin{array}{rcl} \tan (2\alpha +\beta )&=& \tan (\alpha + (\alpha + \beta ))\\ &=& \frac{\tan \alpha + \tan (\alpha + \beta )}{1-\tan \alpha \tan (\alpha + \beta )}\\ &=& \frac{ \frac{5}{x}+\frac{10}{x} }{ 1-(\frac{5}{x})(\frac{10}{x}) }\\ &=& \frac{\frac{15}{x}}{\frac{x^2-50}{x^2}}\\ &=& \frac{15x}{x^2-50} \end{array}$

Range:

Since $2\alpha +\beta$ is acute (1st quadrant), $\tan (2\alpha +\beta )$ is positive.

$x^2-50 >0$

$x>\sqrt{50}$