Question

In the figure, $AL$ is perpendicular to $BCandCM$ is perpendicular to $AB$. If $CL=AL=2BL$, find $MC/AM$.

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

Answer 1

The correct option is B $3$

In triangle ALC, it known that AL = CL with $\angle L={90}^0$
This makes $\Delta ALC$ an isosceles right triangle with $\angle C=\angle CAL={45}^0$.

In $\Delta LAB,let\angle LAB={\vartheta }^0$.

In $\Delta CAM,\angle A=(45+{\vartheta )}^0$. To find $\frac{MC}{AM},wehavetofindtan(45+\vartheta )$.

$tan(45+\vartheta )=\frac{tan45+tan\vartheta }{1-tan45\ast tan\vartheta }=\frac{1+\frac{BL}{AL}}{1-\frac{BL}{AL}}=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3$. [given CL = AL = 2BL]

Hence the ratio of MC to AM is $3$.