# TrigonometryUsing laws of sines and cosines to solve for a triangle

#### Charlotte

##### New member
Hello, I have problem with this task, I must solve the triangel if i know a:b=2:3, c=15 cm, alfa:beta=1:2.Can you help me please?If you know it write me way how you solve it Thank so much.

#### MarkFL

Staff member
I would begin with the Law of Sines and state:

$$\displaystyle \frac{\sin(\theta)}{2x}=\frac{\sin(2\theta)}{3x}$$

Can you show this implies:

$$\displaystyle \cos(\theta)=\frac{3}{4}$$?

#### MarkFL

Staff member

We have:

$$\displaystyle \frac{\sin(\theta)}{2x}=\frac{\sin(2\theta)}{3x}$$

Multiplying through by $$6x$$ and using a double-angle identity we obtain:

$$\displaystyle 3\sin(\theta)=4\sin(\theta)\cos(\theta)$$

As presumably $$\sin(\theta)\ne0$$ (otherwise our triangle is degenerate) we may divide by this quantity to get:

$$\displaystyle 3=4\cos(\theta)\implies \cos(\theta)=\frac{3}{4}$$

And so the 3 interior angles are:

$$\displaystyle \theta=\arccos\left(\frac{3}{4}\right)$$

$$\displaystyle 2\theta=2\arccos\left(\frac{3}{4}\right)$$

$$\displaystyle \pi-3\arccos\left(\frac{3}{4}\right)$$

As these 3 angles are different, we know we have a scalene triangle, and knowing this will help.

Now, let's use the Law of Cosines as follows:

$$\displaystyle (2x)^2=(3x)^2+15^2-2(3x)(15)\cos(\theta)$$

$$\displaystyle 4x^2=9x^2+15^2-90x\left(\frac{3}{4}\right)$$

Arrange resulting quadratic in standard form:

$$\displaystyle 2x^2-27x+90=0$$

Factor:

$$\displaystyle (2x-15)(x-6)=0$$

Because $$2x$$ cannot be 15 as all three sides must have different lengths, we reject that root and are left with:

$$\displaystyle x=6$$

And so, our triangle has:

Side $$a$$ is 12 cm and the angle opposing it is about $$\displaystyle 41.4^{\circ}$$.

Side $$b$$ is 18 cm and the angle opposing it is about $$\displaystyle 82.8^{\circ}$$.

Side $$c$$ is 15 cm and the angle opposing it is about $$55.8^{\circ}$$.