Question

Let $ABC$ be a triangle with $AB=AC$ and $D$ is mid-point of $BC,E$ is the foot of perpendicular drawn from $D$ to $AC$ and $F$ the mid point of $DE$. Angle between the line $AF$ and $BE$ is $\theta$. Then the value of $4\sin \theta$ is

• A. $4$
• B. $3$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $4$

Let the coordinates $A(0,0),B(2b,2c)$ and $C(2a,0)$ as shown in the figure.
$D$ is the mid-point of $BC$, hence $D$ is $(a+b,c)$,
Foot of perpendicular $E$, drawn from $D$ on $AC$ is $E(a+b,0)$
$F$ is the mid-point of $DE$, hence $F$ is $(a+b,\frac{c}{2})$
As $AB$ and $AC$ are equal,
$\therefore a^2=b^2+c^2$
Now,
The slope of $BE$ is: $m_1=\frac{2c}{b-a}$ and
Slope of $AF$ is: $m_2=\frac{c}{2(b+a)}$
$m_1\cdot m_2=\frac{2c}{b-a}\cdot \frac{c}{2(b+a)}=\frac{c^2}{b^2-a^2}=-1$
$\therefore$ Both $BE$ and $AF$ are perpendicular to each other, $\theta ={90}^o$
$\sin \theta =\sin {90}^o=1$
$4\sin \theta =4\sin {90}^o=4$