Apply the angles sum property in the triangle ABC,
[tex]62+90+\angle ACB=180\Rightarrow\angle ACB=180-152=28^{}[/tex]
Similarly, apply the angles sum property in triangle BCD,
[tex]20+90+\angle BCD=180\Rightarrow\angle BCD=180-110=70[/tex]
From triangle ABC,
[tex]BC=AC\sin 62=30\sin 62\approx26.5[/tex]
From triangle BDC,
[tex]BD=BC\cos 20=26.5\cos 20\approx24.9[/tex]
Now, consider that,
[tex]\angle BDE+\angle BDC=180\Rightarrow\angle BDE+90=180\Rightarrow\angle BDE=90[/tex]
So the triangle BDE is also a right triangle, and the trigonometric ratios are applicable.
Solve for 'x' as,
[tex]x=\tan ^{-1}(\frac{BD}{DE})=\tan ^{-1}(\frac{24.9}{16})=57.2764\approx57.3[/tex]
Thus, the value of the angle 'x' is 57.3 degrees approximately.ang
