Using laws of sines and cosines to solve for a triangle

[Charlotte]

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]

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]

To follow up:

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 \(55.8^{\circ}\).

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