Answer:
[tex]x \approx 24\degree [/tex]
Step-by-step explanation:
In triangle BAC, by Pythagoras theorem:
[tex]AC^2 = BC^2 -BA^2 [/tex]
[tex]AC^2 = (15)^2 -(9)^2 [/tex]
[tex]AC^2 = 225 - 81[/tex]
[tex]AC^2 = 144[/tex]
[tex]AC = \sqrt {144} [/tex]
[tex]AC = 12\: cm [/tex]
In triangle ABD,
[tex] \tan \angle ABD =\frac{AD}{AB} [/tex]
[tex] \tan \angle ABD =\frac{5}{9} [/tex]
[tex] \angle ABD =\tan^{-1}\frac{5}{9} [/tex]
[tex] \angle ABD =29.054604099\degree [/tex]
[tex] \angle ABD =29.05\degree [/tex]
Next, in triangle ABC, by sine rule:
[tex] \frac{\sin \angle ABC}{AC}=\frac{\sin \angle BAC}{BC}[/tex]
[tex] \frac{\sin (\angle ABD+\angle DBC) }{12}=\frac{\sin 90\degree}{15}[/tex]
[tex] \frac{\sin (29.05\degree+x) }{12}=\frac{1}{15}[/tex]
[tex] \sin (29.05\degree+x) =\frac{12}{15}[/tex]
[tex]29.05\degree+x = \sin^{-1}\frac{12}{15}[/tex]
[tex]29.05\degree+x =53.130102354\degree [/tex]
[tex]29.05\degree+x =53.13\degree [/tex]
[tex]x =53.13\degree- 29.05\degree[/tex]
[tex]x =24.08\degree [/tex]
[tex]x \approx 24\degree [/tex]
