Question

In the figure given below $AL$ is perpendicular to $BC$ and $CM$ is perpendicular to AB If $CL=AL=2BL$ find $\frac{MC}{AM}$

• A. $2$
• B. $3$
• C. $4$
• D. Cannot be determined

Answer 1

The correct option is B $3$

Let, $BL=l$ $⟹AL=CL=2l$

$AL=CL⟹\angle ACL=\angle CAL={45}^{\circ }$

In $△ACL$, By Pythagoras Theorem,

$A{C}^2=A{L}^2+C{L}^2$

$A{C}^2=2A{L}^2$------(given $AL=CL$)

$AC=\sqrt{2\left(4l^2\right)}=2l\sqrt{2}$

In $△ABL$, By Pythagoras Theorem,

$A{B}^2=A{L}^2+B{L}^2$

$A{B}^2=l^2+(2l{)}^2$

$ABC=\sqrt{5l^2}=l\sqrt{5}$

In $△ABC$, by $Sin$ law

$\frac{BC}{\sin A}=\frac{AB}{\sin C}=\frac{AC}{\sin B}$

$\frac{3l}{\sin A}=\frac{\sqrt{5}l}{\sin C}=\frac{2\sqrt{2}l}{\sin B}$

$\frac{3}{\sin A}=\frac{\sqrt{5}}{\frac{1}{\sqrt{2}}}=\frac{2\sqrt{2}}{\sin B}$

$\frac{3}{\sin A}=\frac{\sqrt{5}}{\frac{1}{\sqrt{2}}}$ and $\frac{\sqrt{5}}{\frac{1}{\sqrt{2}}}=\frac{2\sqrt{2}}{\sin B}$

$\sin A=\frac{3}{\sqrt{10}}$ and $\sin B=\frac{2}{\sqrt{5}}$

If $\sin A=\frac{Opp.side}{Hypotenuse}=\frac{3}{\sqrt{10}}$

then, $\cos A=\frac{Adj.side}{Hypotenuse}=\frac{1}{\sqrt{10}}$

In $△AMC$

$\tan A=\frac{MC}{AM}=\frac{\sin A}{\cos A}=\frac{\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}}$

$\therefore \frac{MC}{AM}=3$