SOLUTION: Find the sides (to 2 decimal places) of the following triangle: angle A = 45°, angle B = 60°, angle C = 75°. In between B and C is 5

Draw CD perpendicular to AB



Triangle BCD is a 30°-60°-90° triangle and so its shorter leg BD is half 
of its hypotenuse BC which is 5, and half of 5 is 2.5. So BD = 2.5

Now we can find CD either by the Pythagorean theorem or from our
knowledge that the longer leg of a 30°-60°-90° right triangle is the 
shorter leg times the square root of 3. Either way you get 
CD = 2.5× = 4.330127019.

Now triangle ACD is a 45°-45°-90° or isosceles right triangle. 
So AD = CD = 4.330127019

Therefore AB = AD + BD = 4.330127019 + 2.5 = 6.830127019.

Now we can find AC either by the Pythagorean theorem or from our
knowledge that the hypotenuse of a 45°-45°-90° triangle is a leg 
times the square root of 2. Either way you get 
AC = AD× = 4.330127019 = 6.123724357.

Rounding off to two decimals:

AB = 6.83
AC = 6.12
BC = 5  (given).

Edwin