In the diagram, ABC is a triangle in which AB = 4 cm, BC = 6 cm and angle ABC = 150°. The line CX is perpendicular to the line ABX. (i) Find the exact length of BX and show that angle CAB = tan^-1 (3/4+3√3)
(ii)
Show that the exact length of AC is √52 + 24√3 cm.

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,