In the given figure AB = DB and AC = DC. If $\angle A B D=58^{\circ}$ and $\angle DBC=(2x-4)^{\circ}, \angle A C B=(y+15)^{\circ}$ and $\angle D C B=63^{\circ}$, then the value of 2x + 5y is :

273

As AB = DB, AC = DC, and BC is common for two triangle

So, \(\Delta \)ABC = \(\Delta \)DBC

So, \(\angle\)ABC = \(\angle\)DBC = \(\angle\)ABD/2

⇒ \(\frac{58}{2}\) = \({29}^\circ\)

So,

\({2x\; -\; 4 }^\circ\) = \({29}^\circ\)

⇒ 2x = \({33}^\circ\)

Again,

\(\angle\)ACB = \(\angle\)DCB = \({63}^\circ\)

So,

(y + \({15}^\circ\)) = \({63}^\circ\)

⇒ y = \({48}^\circ\)

So,

2x + 5y = \({33}^\circ\) + 5 x \({48}^\circ\)

⇒ \({33}^\circ\) + \({240}^\circ\)

⇒ \({273}^\circ\)

Therefore, the answer is \({273}^\circ\)