Question

10. In the given figure,

AB= DB and
AC = DC.
If ZABD = 58°,
B В
ZDBC = (2x - 4),
ZACB = y + 15° and
D
ZDCB = 63°; find the values of x and y.​

Answer 1

there is no fig so i can't answer it

Answer 2

$\sf \: In \: \triangle \: ABC \: and \: \triangle \: BDC, \\ \\ \sf \: AB \: = \: BD \: (given) \\ \sf \: AC \: = \: DC \: (given) \\ \sf \: BC \: = \: BC \: (common) \\ \sf \therefore \triangle \: ABC \cong \triangle \: BDC \: (S.S.S) \\ \sf \therefore \angle \: ABC \: = \: \angle \: DBC \: (C.P.C.T) \\ \\ \sf \: Now, \\ \sf \: \angle \: ABC \: = \: \angle \: DBC \: (proved \: above) \\ \sf \: \angle \: ABC \: = \: 58 \degree \: - \: (2x - 4) \degree \\ \sf \: \angle \: ABC \: = \: 58 \degree \: - \: 2x + 4 \degree \\ \sf \: \angle \: ABC \: = \:( 62 - 2x) \degree \\ \\ \sf \: (62 - 2x) \degree \: = \: (2x - 4) \degree \\ \sf 62 + 4 = 2x + 2x \\ \sf \: 66 = 4x \\ \sf \: x = \frac{66}{4} \\ \boxed{ \red{\sf {\implies \:x = 16.5 \degree}}} \\ \\ \sf \: Now, \\ \sf \: \angle \: BCD \: = \angle \: ACB \: (proved \: above) \\ \sf 63 \degree \: = \: y + 15 \degree \\ \sf \: y \: = \: 63 - 15 \degree \\ \boxed{ \red{ \sf{ \implies{y \: = \: 48 \degree}}}}$