Question

Consider 3 lines
$L_1:5x-y+4=0$
$L_2:3x-y+5=0$
$L_3:x+y+8=0$
If theses lines enclose a triangle ABC and sum of the squares of the tangent of the interior angles can be expressed in the form of $p/q$ where $p$ and $q$ are relatively prime numbers, compute the value of $(p+q)$.

• A. $500$
• B. $450$
• C. $230$
• D. $465$

Answer 1

Answer: (D)

$465$

Given,

${\tan }^2{\theta }_1+{\tan }^2{\theta }_2+{\tan }^2{\theta }_3=\frac{p}{q}$

We have the formula,

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Using this formula,

$\tan {\theta }_2=\left|\frac{5-3}{1+5.3}\right|=\left|\frac{2}{16}\right|=\frac{1}{8}$

$\tan {\theta }_3=\left|\frac{3-(-1)}{1+3.(-1)}\right|=\left|\frac{4}{-2}\right|=2$

$\tan {\theta }_1=\left|\frac{5-(-1)}{1+5.(-1)}\right|=\left|\frac{6}{-4}\right|=\frac{3}{2}$

$\therefore {\tan }^2{\theta }_1+{\tan }^2{\theta }_2+{\tan }^2{\theta }_3=(\frac{3}{2}{)}^2+(\frac{1}{8}{)}^2+2^2=\frac{9}{4}+4+\frac{1}{64}=\frac{144+1+256}{64}=\frac{401}{64}$

$\frac{p}{q}=\frac{401}{64}$

Hence, $p+q=401+64=465$