Question

AD bisects angle CAB. If angle CAD=(8x+6)° and angle DAB=(x+20)°, what is the value of x?​

Answer 1

AnSwEr :

Here, we have to find the value of x,

AD bisects <CAB ,

$\angle\sf CAD \: = (8x + 6) \\ \angle \sf \: DAB = (x + 20)$

• x = ?

Let's take these 2 hint of angles as 2 Equations ;

$\angle \rm \: CAD \: = (8x + 6)..........(1) \\ \angle \rm \: DAB \: = (x + 20)............(2)$

From substracting equation 1 from 2, we get,

$\sf \: - 7x = 14$

$\to \: \sf \: - 7x = 14 \\ \\ \to \sf x = \frac{14}{ - 7} \\ \\ \to\bf \: x = 2$

${\boxed{\frak{x = 2}}} \:$

$\therefore$X = 2°

Let's find the angle :

As AD bisects <CAD , then the 2 angles (CAD and DAB) are equal.

$\angle \sf CAD \: = (8x + 6) \\ \\ \to \sf \: ( 8 \times 2) + 6 \\ \\ \to \sf \: 16 + 6 \\ \\ \bf \to \: 22$

$\angle \sf \: DAB =( x + 20) \\ \\ \to \sf 2 + 20 \\ \\ \to \bf \: 22$

$\angle \sf \: A = 22° + 22° \\ = 44°$

So , the <A = 44° and AD bisects it as 22°.