Question

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

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

Answer 1

The correct option is B $3$

Given,

$CL=AL=2BL$

Let, $BL=l$

$\therefore CL=Al=2l$

In $△ACL$, $AL=CL$ and $\angle ALC={90}^{\circ }$

$\therefore \angle ACL=\angle CAL={45}^{\circ }$

By Pythagoras theorem,

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

$A{C}^2=(2l{)}^2+(2l{)}^2=8l^2$

$AC=2\sqrt{2}l$

In $△ABL$,

By Pythagoras theorem,

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

$A{B}^2=(2l{)}^2+(l{)}^2=5l^2$

$AC=\sqrt{5}l$

In $△ABC$, by $\sin$ rule,

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

$BC=BL+CL$

$\frac{3l}{\sin A}=\frac{2\sqrt{2}l}{\sin B}=\frac{\sqrt{5}l}{\sin {45}^{\circ }}$

$\frac{3l}{\sin A}=\frac{2\sqrt{2}l}{\sin B}=\frac{\sqrt{5}l}{\sin {45}^{\circ }}$

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

$\sin A=\frac{3}{\sqrt{10}}$ and $\sin B=\frac{2}{\sqrt{5}}$ ----------(1)

In $△AMC$,

$\sin A=\frac{CM}{AC}=\frac{3}{\sqrt{10}}$

By Pythagoras theorem,

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

$10=A{M}^2+9$

$A{M}^2=10-9$

$AM=1$

$\tan A=\frac{CM}{AM}=\frac{3}{1}$

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