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
if you're curious as to how that looks on paper.
Using your calculator, you should find that,
