AB (240 m) is a base of an object that is tilted over the ramp with angle BCE = 45 degrees. CE is 60 m, ie. the known section of the object on the ground say if ramp had not been there.
My aim is to find out the value of BAC, ie the angle at which the object is tilted.
I have tried and could find out the length of various segments. But can't get to any way that can relate it with the angle BAC
By the law of cosines$$AB^2=AC^2+BC^2-2AC\cdot BC\cos135^o$$So$$BC^2=AB^2-AC^2+2AC\cdot BC\cos135^o$$or$$BC^2=240^2-180^2-1.414\cdot 180BC$$i.e.$$BC^2=25200-254.52BC$$and$$BC^2+254.52BC-25200=0$$By the quadratic formula:$$BC=\frac{-254.52+\sqrt{254.52^2-4\cdot 25200}}{2}$$and$$BC\approx 76.2$$Again by the cosine law$$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2AB\cdot AC}\approx .97$$and hence$$\angle BAC\approx 12.95^o$$