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

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

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JeanaShupp JeanaShupp · Answer 2 · 2019-10-23T08:21:52+02:00

Answer: [tex]\angle{DBF}=115^{\circ}[/tex]

Step-by-step explanation:

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

Segment CD bisects angle ACB.

i.e. [tex]\angle{BCD}\cong\angle{ACD}[/tex]

i.e. [tex]3x-2=5x-20\\\\\Rightarrow\ 5x-3x=20-2\\\\\Rightarrow\ 2x=18\\\\\Rightarrow\ x=\dfrac{18}{2}=9[/tex]

Now, [tex]\angle{BCD}=(5(9)-20)^{\circ}=25^{\circ}[/tex]

Since in a triangle ,

Exterior angle = Sum of opposite interior angles

i.e. [tex]\angle{DBF}=\angle{BCD}+\angle{CDB}[/tex]

[tex]\angle{DBF}=25^{\circ}+90^{\circ}=115^{\circ}[/tex]

Hence, [tex]\angle{DBF}=115^{\circ}[/tex]