I forgot my trig but I can see if I can do something this will be a confusing solution
if we figure out the 3rd angle, it is 55 therefor we has a sideways isocoleese triangle (see attachment) we can draw an auxilary (imaginary) line from left tower to left side to make 2 identical right triangles with hypotonuse 5 and the angles are 35 and 55
erase all that, and now we have draw an immaginary line (see attachmen 2) from bottom to top to represent height I am going to try to find the base (leg) cos=a/h cos(70°)=x/5 times both sides by 5 5cos(70°)=x (from left tower to perpendicular line) now
a²+b²=c² (5cos(70°)²+h²=5² 25(cos(70°))²+h²=25 minus 25(cos(70°))² both sides h²=25-25(cos(70°))² h²=25(1-(cos(70°)²)) square root both sides h=5√(1-(cos(70°)²)) or 5cos(pi/9) evaluate with your calculator h=4.6984km round h=4.7km
