# Thread: Triangle Trouble

1. ## Triangle Trouble

In a triangle ABC, X is a point on AC such that AX/XC=3 and the angle AXB = 70 degrees. Find the length of AC if the angle of ABC = 110 degrees and BC = 6 centimetres.

thanks so much

2. By Similar Triangles,

Angle ABC = Angle BXC
Angle ACB = Angle XCB
Angle BAC = Angle XBC

XC=1/4*AC

Solve for AC in AC/6=6/XC =$\displaystyle \sqrt{144}$

Hope that helps.

3. Originally Posted by googa
In a triangle ABC, X is a point on AC such that AX/XC=3 and the angle AXB = 70 degrees. Find the length of AC if the angle of ABC = 110 degrees and BC = 6 centimetres.

thanks so much
In straight angle AXC,
angle BXC = 180 -70 = 110 degrees

In triangle BXC,
Let angle XBC = theta
(theta +angle C) = 180 -110 = 70 deg

In big triangle ABC
(angle A +angle C) = 180 -110 = 70 deg also

So,
theta +angle C = angle A +angle C
Hence,
angle A = theta

---------------
If AX /XC = 3,
then side AC is divided into 4 equal parts.
Let w = length of AC,
So, XC = w/4

In triangle BXC, by Law of Sines,
BC /sin(angle BXC) = XC /sin(angle XBC)
6 /sin(110deg) = (w/4) /sin(theta)
Cross multiply,
6sin(theta) = (w/4)sin(110deg)
sin(theta) = w*sin(110deg) /24 ------**

In triangle ABC, by Law of Sines,
AC /sin(angle ABC) = BC /sin(angle A)
w /sin(110deg) = 6 /sin(theta)
Cross multiply,
w*sin(theta) = 6sin(110deg)
sin(theta) = 6sin(110deg) /w -----**

sin(theta) = sin(theta)
w*sin(110deg) /24 = 6sin(110deg) /w
w/24 = 6/w
w^2 = 6*24
w = sqrt(144) = 12

Therefore, AC = 12 cm ----------answer.

4. Hello, googa!

A slight variation of ticbol's solution . . .

In $\displaystyle \Delta ABC,\:X$ is a point on $\displaystyle AC$ such that $\displaystyle \frac{AX}{XC}=3$

and $\displaystyle \angle AXB = 70^o,\;\;\angle ABC = 110^o,\;\;BC = 6$ cm.

Find the length of $\displaystyle AC.$
Code:
                              B
o
*   **
*       * *
*           * β*  6
*               *   *
*                   *    *
*                       *     *
* β                     70° *    α *
o   *   *   *   *   *   *   *   o   *   o
A               3a              X   a   C

We have: .$\displaystyle \angle ABC = 110^o,\;\;\angle AXB = 70^o,\;\;BC = 6$
Let: $\displaystyle a = XC,\;AX = 3a$

Consider $\displaystyle \Delta ABC$ and $\displaystyle \Delta BXC$

. . They share $\displaystyle \angle \alpha = \angle BCX$

Since $\displaystyle \angle AXB = 70^o$, then $\displaystyle \angle BXC = 110^o$

. . They have equal angles: .$\displaystyle \angle ABC \:=\:\angle BXC \:=\:110^o$

Hence, their third angles are equal: .$\displaystyle \angle BAC \:=\:\angle CBX \:=\:\beta$

In $\displaystyle \Delta ABC\!:\;\;\frac{\sin\beta}{6} \:=\:\frac{\sin110^o}{4a} \quad\Rightarrow\quad \sin\beta \:=\:\frac{3\sin110^o}{2a}$ .[1]

In $\displaystyle \Delta BXC\!:\;\;\frac{\sin\beta}{a} \:=\:\frac{\sin110^o}{6} \quad\Rightarrow\quad \sin\beta \:=\:\frac{a\sin110^o}{6}$ .[2]

Equate [1] and [2]: .$\displaystyle \frac{3\sin110^o}{2a} \:=\:\frac{a\sin110^o}{6} \quad\Rightarrow\quad 18 \:=\:2a^2 \quad\Rightarrow\quad a \:=\:3$

Therefore: .$\displaystyle AC \:=\:4a \:=\:\boxed{12\text{ cm}}$