Question 551254
find the sides (to2 decimal places) of the following triangle:
angle A = 45, angle B = 60, angle C = 75. In between B and C is 5 
<pre>
{{{drawing(400,275,-.5,7.5,-.5,5, locate(5.75,2.2,5),
locate(.4,.35,"45°"),locate(0,0,A), locate(6.8,0,B),locate(4.3,4.7,C),
triangle(0,0,2.5sqrt(3)+2.5,0,2.5sqrt(3),2.5sqrt(3)),locate(6.1,.35,"60°"),
locate(4,4,"75°") )}}}
Draw CD perpendicular to AB

{{{drawing(400,275,-.5,7.5,-.5,5, locate(5.75,2.2,5),
locate(.4,.35,"45°"),locate(0,0,A), locate(6.8,0,B),locate(4.3,4.7,C),
triangle(0,0,2.5sqrt(3)+2.5,0,2.5sqrt(3),2.5sqrt(3)),locate(6.1,.35,"60°"),
locate(4,4,"75°"), green(line(2.5sqrt(3),2.5sqrt(3),2.5sqrt(3),0), locate(4.3,0,D))
 )}}}

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×{{{sqrt(3)}}} = 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×{{{sqrt(2)}}} = 4.330127019{{{sqrt(2)}}} = 6.123724357.

Rounding off to two decimals:

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

Edwin</pre>