Part (i)
Angle ABC is 150 degrees. The angle CBX is supplementary to this, so,
(angleABC)+(angleCBX) = 180
angle CBX = 180-(angleABC)
angle CBX = 180-150
angle CBX = 30 degrees
Because of the right angle marker at point X, we can find that triangle CBX is a 30-60-90 triangle. This special type of right triangle has its hypotenuse twice as long as the short leg, which we'll call y. The long leg is equal to y*sqrt(3) units.
The diagram shows BC = 6 is the hypotenuse, so the short leg is y = 6/2 = 3. The longer leg is y*sqrt(3) = 3*sqrt(3) which is the distance from B to X.
In summary so far,
CX = 3
BX = 3*sqrt(3)
We have enough information to find the tangent of angle CAB, and then find the actual angle itself.
tan(angle CAB) = opposite/adjacent
tan(angle CAB) = (CX)/(AX)
tan(angle CAB) = 3/(4+3*sqrt(3))
angle CAB = arctan( 3/(4+3*sqrt(3)) )
I'm using arctan in place of tan^(-1) which is the same basic function, just different notation.
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Part (ii)
Focus on triangle AXC. The legs we found were
AX = 4+3*sqrt(3)
CX = 3
Use the Pythagorean theorem to find the hypotenuse AC
a^2 + b^2 = c^2
(AX)^2 + (CX)^2 = (AC)^2
(AC)^2 = (AX)^2 + (CX)^2
(AC)^2 = (4+3*sqrt(3))^2 + (3)^2
(AC)^2 = (4+3*sqrt(3))(4+3*sqrt(3)) + 9
(AC)^2 = 4(4+3*sqrt(3))+3*sqrt(3)(4+3*sqrt(3)) + 9
(AC)^2 = 16+12*sqrt(3)+12*sqrt(3)+27 + 9
(AC)^2 = 52+24*sqrt(3)
AC = sqrt( 52+24*sqrt(3) )
Note that AC is a length so AC is positive. We don't have to worry about the plus minus.
This is known as a nested radical since one square root is buried in another.
We can write that as [tex]AC = \sqrt{52 + 24\sqrt{3}}[/tex] if you're curious as to how that looks on paper.
Using your calculator, you should find that,
[tex]\sqrt{52+24\sqrt{3}} \approx 9.673118[/tex]
