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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

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    $\begingroup$ @Vishwas Have you tried to use the law of sines and/or the law of cosines? $\endgroup$
    – John Omielan
    Commented Jul 22, 2023 at 19:52
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    $\begingroup$ Hint: You don't need $E$. You have $AB = 240, AC = 180, ACB = 135°$ $\endgroup$
    – Gribouillis
    Commented Jul 22, 2023 at 20:05
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    $\begingroup$ For more details on how to apply the laws of sines or cosines, see solution of triangles on Wikipedia. $\endgroup$
    – peterwhy
    Commented Jul 22, 2023 at 20:18
  • $\begingroup$ @Gribouillis I don't want ACB . I want value of BAC. $\endgroup$
    – Vishwas
    Commented Jul 23, 2023 at 4:16
  • $\begingroup$ @peterwhy I know the formulae quite well. But this one proved to be difficult to sort out. $\endgroup$
    – Vishwas
    Commented Jul 23, 2023 at 4:16

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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$$\angle BAC\approx 12.95^o$$

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