Question

please solve Questions no.6 please

Answer 1

$\huge\mathcal{Heya}$

$\mathsf\red{Given \ :-}$

∆ ABC and ∆ DBC are isosceles triangles.

/_BDC = 80°

/_BAC = 30°

$\mathsf\red{To \ find \ :-}$

(a) /_ABC and /_ACB

(b) /_DBC and /_DCB

(c) /_x

$\mathsf\red{Solution \ :-}$

In ∆ ABC,

AB = AC (As two sides are equal in isosceles triangle.)

•°• /_ABC = /_ACB ______(1) (Angles opposite to equal sides are equal.)

Now,

/_ABC + /_ACB + /_BAC = 180° (Angle Sum Property)

/_ABC + /_ABC + /_BAC = 180° [From (1)]

2/_ABC + 30° = 180°

2/_ABC = 150°

/_ABC = 75°

So,

/_ABC = /_ACB = 75° [From (1)]

In ∆ DBC,

DB = DC (As two sides are equal in isosceles triangle.)

•°• /_DBC = /_DCB ______(2) (Angles opposite to equal sides are equal.)

Now,

/_DBC + /_DCB + /_BDC = 180° (Angle Sum Property)

/_DBC + /_DBC + /_BDC = 180° [From (1)]

2/_DBC + 80° = 180°

2/_DBC = 100°

/_DBC = 50°

So,

/_DBC = /_DCB = 50° [From (2)]

Now,

x = /_ABC - /_DBC

Putting values

x = 75° - 50°

x = 25°

Hence,
$<b>$
(a) /_ABC = /_ACB = 75°

(b) /_DBC = /_DCB = 50°

(c) /_x = 25°

$\huge\mathbb{Hope \ this \ helps.}$