Answer:
[tex]x = 10[/tex]
Step-by-step explanation:
The attachment completes the question;
From the question, we have the following given parameters
[tex]BD = CD[/tex]
[tex]\angle BCD = 15\deg[/tex]
[tex]\angle BAD = 5\deg[/tex]
Required
Solve for x
Since [tex]BD = CD[/tex] then
[tex]\angle CBD = \angle BCD[/tex] --- Property: Base angle of an isosceles triangle
[tex]\angle CBD = \angle BCD = 15[/tex]
Also, ABC is a straight line;
So:
[tex]\angle CBD + \angle DBA = 180[/tex] --- angle on a straight line
Substitute 15 for [tex]\angle CBD[/tex]
[tex]15 + \angle DBA = 180[/tex]
Make [tex]\angle DBA[/tex] the subject
[tex]\angle DBA = 180 - 15[/tex]
[tex]\angle DBA = 165[/tex]
Considering triangle DBA
[tex]\angle DBA + \angle BAD + x = 180[/tex] ---- angles in a triangle
Substitute 165 for [tex]\angle DBA[/tex] and 5 for [tex]\angle BAD[/tex]
[tex]165 + 5 + x = 180[/tex]
Make x the subject
[tex]x = 180 - 165 -5[/tex]
[tex]x = 10[/tex]
