SOLUTION: ABC is a triangle with ∠CAB=15 ∘ and ∠ABC=30 ∘ . If M is the midpoint of AB, Sin then ∠ACM= ?

  
This problem would be much easier if you could use decimals, but 
decimals are approximate. The answer could only be approximate if we 
use decimals, so we must use exact values.  Maybe some other tutor can 
come up with a shorter solution, but this was the very long solution I 
came up with:



We want to find angle θ 

Without loss of generality we can let AM = MB = 1 unit.

We are going to need expressions for both sine and cosine of 15o.





Angle ACB = 180o-15o-30o = 135o.

By the law of sines on triangle ABC.
























By the law of cosines on triangle ACM













Now let's see if we can rewrite that so it doesn't have a 
square root under a square root.  We'll see if we can write
it as the difference of 2 square roots:



=>



 =>  => 
CM can't be negative, so 



Use the law of sines on triangle ACM:







Edwin

The given part is shown in Figure 1 below.




                       Figure 1.


Triangle ABC has the angles  A= 15°,  B= 30°  and  C= 180°-15°-30°= 135°.

We want to find angle .


Draw perpendicular CH from vertex C to the base AB (Figure 2).




                       Figure 2.


First part of the solution is the same as that of Edwin' solution,

and it shows that a = , b = .


The rest of the solution is different (very simple and totally geometrical).


Triangle BHC is a right-angled triangle with the acute angle B of 30°.
Hence, the opposite leg CH is half of the hypotenuse BC

    CH =  = ,    (1)


while its other leg BH is    times the hypotenuse BC

    BH =  = .


Then the segment MH is the complement of BH to 1

    MH = 1 -  =  = .    (2)


Comparing expressions (1) and (2), we see that CH = MH.


Hence, right-angled triangle MHC is isosceles right-angled triangle,
which implies that angle MCH is 45°.


Thus  ∠ACM = 135° - 60° - 45° = 30°,


and the problem is solved completely.