.
Let's let /_BDC= /_x and /_DBC= /_y
Since triangle ABC is an isosceles triangle then /_ACB=80^@
Then from triangle BDC we have:
(BC)/sinx=(BD)/sin80=(DC)/siny
From triangle ABC we have:
(AC)/sin80=(BC)/sin20
But AC=AD+DC and BC=AD. Therefore:
AC=BC+DC
(BC+DC)/sin80=(BC)/sin20
BCsin20+DCsin20=BCsin80
DCsin20=BC(sin80-sin20)
DC=(BC(sin80-sin20))/sin20
(BC)/sinx=((BC(sin80-sin20))/sin20)/siny
(BC)/sinx=(BC(sin80-sin20))/(sinysin20)
(cancel color(red)(BC))/sinx=(cancelcolor(red)(BC)(sin80-sin20))/(sinysin20)
sinysin20=sinx(sin80-sin20)
From triangle BCD, we know x+y=100^@
y=100-x
sin(100-x)sin20=sinx(sin80-sin20)
Using the trigonometric identity:
sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta
(sin100cosx-cos100sinx)sin20=sinx(sin80-sin20)
sin20sin100cosx-sin20cos100sinx=sinx(sin80-sin20)
sin20sin100cosx=sinx(sin80-sin20+sin20cos100)
Dividing both sides by cosx
sin20sin100=tanx(sin80-sin20+sin20cos100)
Solving for tanx, we get:
tanx=(sin20sin100)/(sin80-sin20+sin20cos100)
Now, we can plug in the values of sin and cos for each angle and get a value for tanx:
tanx=((0.342)(0.9848))/((0.9848-0.342+(0.342)(-0.1736))
tanx=0.3368/0.5834=0.5773
x=arctan0.5773=29.9978^@ or approximately 30^@
/_ADB=180-/_BDC=180-/_x=180-30=150^@