Question

Perpendiculars are drawn from angles $A,B,C$ of an acute angled triangle on the opposite sides and produced to meet the circumscribing circle. If these produced parts be $\alpha ,\beta ,\gamma$ respectively, then the value of $\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }$ is equal to

• A. $\tan A+\tan B-\tan C$
• B. $2(\tan A+\tan B+\tan C)$
• C. $\tan A+\tan B+\tan C$
• D. $2(\tan A-\tan B+\tan C)$

Answer 1

The correct option is B $2(\tan A+\tan B+\tan C)$

Let $AD$ be perpendicular from $A$ on $BC$ when $AD$ is produced, it meets the circumscribing circle at $E$ from question $DE=\alpha$
Since, angles in the same segment are equal
$\angle AEB=\angle ACB=\angle C$ and $\angle AEC=\angle BC=\angle B$
From the right angled triangle $BDE$
$\tan C=\frac{BD}{DE}$ ... (i)
From the right angled triangle $CDE$
$\tan B=\frac{CD}{DE}$ ... (ii)
Adding (i) and (ii), we get
$\tan B+\tan C=\frac{BD+CD}{DE}=\frac{BC}{DE}=\frac{a}{\alpha }$ ...(iii)
Similarly $\tan C+\tan A=\frac{b}{\beta }$ ... (iv)
$\tan A+\tan B=\frac{c}{\gamma }$ ..(v)
And adding (iii), (iv), (v), we get
$\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=2(\tan A+\tan B+\tan C)$