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

• MHB
• Charlotte
In summary, to solve the given triangle with known ratios and angles, the Law of Sines and Law of Cosines were used. The resulting angles and side lengths were found to be 41.4^{\circ} for angle $$a$$, 82.8^{\circ} for angle $$b$$, 55.8^{\circ} for angle $$c$$, and side lengths of 12 cm, 18 cm, and 15 cm respectively.
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.

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:

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

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

#### Attachments

• mhb_0011.png
4.4 KB · Views: 64

## 1. How do you use the law of sines to solve for a triangle?

The law of sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. To use this law to solve for a triangle, you need to know at least two angles and one side length. Then, you can use the formula: sin(A)/a = sin(B)/b = sin(C)/c, where A, B, and C are the angles and a, b, and c are the corresponding side lengths.

## 2. When should I use the law of cosines to solve for a triangle?

The law of cosines is used to solve for a triangle when you know all three side lengths or two side lengths and the included angle. This law states that the square of a side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of those sides and the cosine of the included angle. The formula is c² = a² + b² - 2abcos(C), where c is the side opposite the angle C.

## 3. Can the law of sines and cosines be used to solve any triangle?

Yes, the law of sines and cosines can be used to solve any triangle, as long as you have enough information about the triangle. The law of sines can be used when you know at least two angles and one side length, while the law of cosines can be used when you know all three side lengths or two side lengths and the included angle.

## 4. What is the difference between the law of sines and the law of cosines?

The main difference between the law of sines and the law of cosines is the type of information they require to solve for a triangle. The law of sines uses angles and side lengths, while the law of cosines uses only side lengths or two side lengths and the included angle. Additionally, the law of sines can be used to find any missing angle or side length, while the law of cosines can only be used to find missing side lengths.

## 5. Can the law of sines and cosines be used to solve for non-right triangles?

Yes, the law of sines and cosines can be used to solve for non-right triangles. In fact, they are commonly used to solve for any type of triangle, as long as you have enough information about the triangle. However, the Pythagorean theorem is only applicable to right triangles, so it cannot be used to solve for non-right triangles.

• General Math
Replies
2
Views
1K
• General Math
Replies
1
Views
742
• General Math
Replies
1
Views
817
• General Math
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
429
• General Math
Replies
1
Views
1K
• General Math
Replies
2
Views
970
• General Math
Replies
3
Views
2K
• General Math
Replies
4
Views
2K
• General Math
Replies
4
Views
1K