calcue the R

Geometry Level 2

$ABC$ is a right triangle with $\angle B = 90^\circ$ . $M$ is the midpoint of $AB$ , $\angle ACM = \alpha$ , and $\angle BMC = \beta$ . Find

$\frac {\sin (\beta-\alpha) \cos \beta}{\sin \alpha}$

The answer is 1.

1 solution

Chew-Seong Cheong
Dec 16, 2019

Note that $\angle A = \beta - \alpha$ . Applying sine rule in $\triangle ACM$ , we have $\dfrac {\sin (\beta - \alpha)}{\sin \alpha} = \dfrac {CM}{AM}$ . Then $\dfrac {\sin (\beta-\alpha)\blue{\cos \beta}}{\sin \alpha} = \dfrac {CM}{AM} \times \blue{\dfrac {BM}{CM}} = \boxed 1$ , since $AM=BM$ .

calcue the R

The answer is 1.

1 solution

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