AD bisects angle CAB. If angle CAD=(8x+6)° and angle DAB=(x+20)°, what is the value of x?
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AD bisects angle CAB. If angle CAD=(8x+6)° and angle DAB=(x+20)°, what is the value of x?
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AnSwEr :
Here, we have to find the value of x,
AD bisects <CAB ,
[tex]\angle\sf CAD \: = (8x + 6) \\ \angle \sf \: DAB = (x + 20)[/tex]
Let's take these 2 hint of angles as 2 Equations ;
[tex]\angle \rm \: CAD \: = (8x + 6)..........(1) \\ \angle \rm \: DAB \: = (x + 20)............(2)[/tex]
From substracting equation 1 from 2, we get,
[tex]\sf \: - 7x = 14[/tex]
[tex]\to \: \sf \: - 7x = 14 \\ \\ \to \sf x = \frac{14}{ - 7} \\ \\ \to\bf \: x = 2[/tex]
[tex]{\boxed{\frak{x = 2}}} \: [/tex]
[tex]\therefore[/tex]X = 2°
Let's find the angle :
As AD bisects <CAD , then the 2 angles (CAD and DAB) are equal.
[tex]\angle \sf CAD \: = (8x + 6) \\ \\ \to \sf \: ( 8 \times 2) + 6 \\ \\ \to \sf \: 16 + 6 \\ \\ \bf \to \: 22[/tex]
[tex]\angle \sf \: DAB =( x + 20) \\ \\ \to \sf 2 + 20 \\ \\ \to \bf \: 22[/tex]
[tex]\angle \sf \: A = 22° + 22° \\ = 44°[/tex]
So , the <A = 44° and AD bisects it as 22°.