Question

In ΔBCD, \overline{BD}

BD
is extended through point D to point E, \text{m}\angle BCD = (2x-1)^{\circ}m∠BCD=(2x−1)
∘
, \text{m}\angle CDE = (7x-19)^{\circ}m∠CDE=(7x−19)
∘
, and \text{m}\angle DBC = (x+10)^{\circ}m∠DBC=(x+10)
∘
. Find \text{m}\angle CDE.m∠CDE.

Answer 1

30degrees

Explanation:

Given

Exterior angle m<CDE = 7x - 19 degrees

interior angles are

m<BCD = 2x - 1

m<DBC = x+10

Since the sum of the interior angles is equal to the exterior, hence;

2x - 1 + x+10 = 7x - 19

3x + 9 = 7x - 19

3x - 7x = -19 - 9

-4x = -28

x = 28/4

x = 7

Get m<CDE

m<CDE = 7x - 19

m<CDE = 7(7) - 19

m<CDE = 49 - 19

m<CDE = 30degrees