Question

In the following figure CB is perpendicular to AB, and CD bisects angle ACB. Find angle DBF

Answer 1

Measure angle ACD = 3x - 2
measure angle BCD = 5x - 20
measure angle ACD = measure angle BCD
3x - 2 = 5x - 20
-2 = 2x - 20
18 = 2x
9 = x
x = 9
measure angle BCD = 5x - 20
measure angle BCD = 5(9) - 20
measure angle BCD = 45 - 20
measure angle BCD = 25
measure angle CDA = 90
measure angle BDC = measure angle CDA
measure angle BDC = 90
measure angle DBF = measure angle BCD + measure angle BDC
measure angle DBF = 25 + 90
measure angle DBF = 115

Answer 2

Answer: $\angle{DBF}=115^{\circ}$

Step-by-step explanation:

In the given picture , We are given Δ ABC in which $\angle{BCD}=(3x-2)^{\circ},\ \angle{ACD}=(5x-20)^{\circ}$

Segment CD bisects angle ACB.

i.e. $\angle{BCD}\cong\angle{ACD}$

i.e. $3x-2=5x-20\\\\\Rightarrow\ 5x-3x=20-2\\\\\Rightarrow\ 2x=18\\\\\Rightarrow\ x=\dfrac{18}{2}=9$

Now, $\angle{BCD}=(5(9)-20)^{\circ}=25^{\circ}$

Since in a triangle ,

Exterior angle = Sum of opposite interior angles

i.e. $\angle{DBF}=\angle{BCD}+\angle{CDB}$

$\angle{DBF}=25^{\circ}+90^{\circ}=115^{\circ}$

Hence, $\angle{DBF}=115^{\circ}$