Let ABC be a triangle with A(-3,1) and ACB=, 0

Question

Let $ABC$ be a triangle with $A(-3,1)$ and $\angle ACB=	heta,\ 0<	heta<\dfrac{\pi}2$. If the equation of the median through $B$ is $2x+y-3=0$ and the equation of angle bisector of $C$ is $7x-4y-1=0$, then $	an	heta$ is equal to:

Accepted Answer

$$\therefore$$ $$M\left( {{{a - 3} \over 2},{{b + 1} \over 2}} \right)$$ lies on 2x + y - 3 = 0

$$\Rightarrow$$ 2a + b = 11 ...........(i)

$$\because$$ C lies on 7x - 4y = 1

$$\Rightarrow$$ 7a - 4b = 1 ......... (ii)

$$\therefore$$ by (i) and (ii) : a = 3, b = 5

$$\Rightarrow$$ C(3, 5)

$$\therefore$$ mAC = 2/3

Also, mCD = 7/4

$$ \Rightarrow \tan {\theta  \over 2} = \left| {{{{2 \over 3} - {4 \over 4}} \over {1 + {{14} \over {12}}}}} \right| \Rightarrow \tan {\theta  \over 2} = {1 \over 2}$$

$$ \Rightarrow \tan \theta  = {{2.{1 \over 2}} \over {1 - {1 \over 4}}} = {4 \over 3}$$

Solutions · 1