Lesson An upper level Geometry problem on special (15°,30°,135°)-triangle

An upper level Geometry problem on special (15°,30°,135°)-triangle

Problem 1

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 .



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

We need this well known expression for sine 15°

    sin(15°) = sin(45°-30°) = sin(45°)cos(30°)-cos(45°)sin(30°)}}} =  =  = .    (1)


By the law of sines on triangle ABC.

     = ,

     = ,

     = ,

     = ,

    b = .


Also,

     = ,

     = ,

     = ,

     = ,

     = ,

     = ,

     = ,

    a = .


In what follows, we will use (2) and will not use (3).
So, (3) is derived here for completeness, only.


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 the 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 =  = ,    (4)


while its other leg BH is    times the hypotenuse BC

    BH =  = .


Then the segment MH is the complement of BH to 1

    MH = 1 -  =  = .    (5)


Comparing expressions (4) and (5), 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°.


Hence, sin(∠ACM) = sin(30°) = 1/2,  and the problem is solved completely.