- #1
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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.
Using laws of sines and cosines to solve for a triangle
Answers and Replies
- #2
MarkFL
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MHB
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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:
\(\displaystyle \cos(\theta)=\frac{3}{4}\)?
\(\displaystyle \frac{\sin(\theta)}{2x}=\frac{\sin(2\theta)}{3x}\)
Can you show this implies:
\(\displaystyle \cos(\theta)=\frac{3}{4}\)?
- #3
MarkFL
Gold Member
MHB
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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 \(55.8^{\circ}\).
View attachment 9699
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 \(55.8^{\circ}\).
View attachment 9699
