Law of Sines and Cosines Notes and Worksheets - Lindsay Bowden - Free Printable

Here are the solutions to fill in the blanks and solve the example problems on your worksheet.

Part 1: Filling in the Blanks

DEFINITION
The law of cosines is a way to find missing sides and angles in an oblique (not right) triangle.

WHEN TO USE
* Left Box (SAS): You want to find: side or angle
*(Note: Usually, with SAS you find the missing side first, then use Law of Sines for the angles, but the Law of Cosines can technically find either depending on how you rearrange it. In this specific diagram, you are given two sides and the angle between them, so you typically find the third side.)*
* Right Box (SSS): You want to find: angle
*(Note: When you have all three sides, you use the Law of Cosines to find any of the missing angles.)*

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Part 2: Examples

Example 1: Find the missing side $x$

Step 1: Identify what we know.
We have two sides ($7$ and $5$) and the angle between them ($58^\circ$). We need to find the opposite side, labeled $x$. This is an SAS case.

Step 2: Choose the correct formula.
Since we are looking for side $x$, we use the version of the formula that solves for that side squared:
$$x^2 = 7^2 + 5^2 - 2(7)(5)\cos(58^\circ)$$

Step 3: Calculate the squares and the product.
$$x^2 = 49 + 25 - 70\cos(58^\circ)$$
$$x^2 = 74 - 70\cos(58^\circ)$$

Step 4: Find the cosine value and solve.
Using a calculator, $\cos(58^\circ) \approx 0.5299$.
$$x^2 = 74 - 70(0.5299)$$
$$x^2 = 74 - 37.093$$
$$x^2 = 36.907$$

Step 5: Take the square root.
$$x = \sqrt{36.907}$$
$$x \approx 6.075$$

Rounding to one decimal place, $x \approx 6.1$.

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Example 2: Find the missing angle $x^\circ$

Step 1: Identify what we know.
We have all three sides: $9$, $7$, and the bottom side is unknown? Wait, looking closely at the image for #2:
- Side adjacent to angle $x$ is $9$.
- Side opposite to angle $x$ is $7$.
- The other side adjacent to angle $x$ is not labeled with a number, but there is a $45^\circ$ angle.
Let me re-examine image #2 carefully.
- Left side: unlabeled length.
- Bottom side: $9$.
- Right side: $7$.
- Angle bottom-left: $x^\circ$.
- Angle bottom-right: $45^\circ$.
- Top angle: unlabeled.

Actually, looking at standard problems of this type, usually, you are given SSS or SAS.
In problem #2, we are given:
- Side $c = 9$ (bottom)
- Side $a = 7$ (right)
- Angle $B = 45^\circ$ (bottom right)? No, the $45^\circ$ is inside the triangle at the bottom right vertex.
- We need to find angle $x$ at the bottom left.

This is actually an AAS or ASA situation if we assume the third side is unknown, which would require the Law of Sines, not Cosines. However, the worksheet is titled "Law of Cosines". Let's look closer. Is the side labeled $7$ opposite the $45^\circ$? No, it's adjacent. Is the side labeled $9$ adjacent to $x$? Yes.

Let's re-read the diagram for #2.
- Vertex A (bottom left): Angle $x$.
- Vertex B (bottom right): Angle $45^\circ$.
- Side AC (bottom): Length $9$.
- Side BC (right): Length $7$.
- Side AB (left): Unknown.

If we use the Law of Cosines here, we can find the third side (AB) first because we have SAS (Side-Angle-Side) relative to the top angle? No, we don't have the top angle.
We have Side-Side-Angle (SSA) if we look at angle $45^\circ$, side $9$, and side $7$. Or we can view it as knowing two sides ($9$ and $7$) and the included angle is NOT given. The angle $45^\circ$ is between side $9$ and the unknown side. The angle $x$ is between side $9$ and the unknown side.

Wait, let's look at the position of the numbers again.
- The $7$ is on the right side.
- The $9$ is on the bottom side.
- The angle $45^\circ$ is between the side labeled $7$ and the side labeled $9$.
Correction: The arc for $45^\circ$ is clearly between the side of length $7$ and the side of length $9$.
Therefore, we have SAS: Side $7$, Included Angle $45^\circ$, Side $9$.
We want to find angle $x$. Angle $x$ is opposite the side of length $7$.

To find angle $x$, we must first find the third side (let's call it $y$, the left side) using the Law of Cosines. Then we can use the Law of Cosines again (or Sines) to find $x$. Since this is a Law of Cosines worksheet, let's stick to that.

Step 1: Find the third side ($y$).
The side $y$ is opposite the $45^\circ$ angle.
$$y^2 = 7^2 + 9^2 - 2(7)(9)\cos(45^\circ)$$
$$y^2 = 49 + 81 - 126\cos(45^\circ)$$
$$y^2 = 130 - 126(0.7071)$$
$$y^2 = 130 - 89.095$$
$$y^2 = 40.905$$
$$y = \sqrt{40.905} \approx 6.396$$

So, the left side is approximately $6.396$.

Step 2: Find angle $x$.
Now we have all three sides: $7$, $9$, and $6.396$.
Angle $x$ is opposite the side of length $7$.
Formula: $a^2 = b^2 + c^2 - 2bc\cos(A)$
Here, $7^2 = 9^2 + (6.396)^2 - 2(9)(6.396)\cos(x)$

$$49 = 81 + 40.905 - 115.128\cos(x)$$
$$49 = 121.905 - 115.128\cos(x)$$

Subtract $121.905$ from both sides:
$$49 - 121.905 = -115.128\cos(x)$$
$$-72.905 = -115.128\cos(x)$$

Divide by $-115.128$:
$$\cos(x) = \frac{-72.905}{-115.128}$$
$$\cos(x) \approx 0.63325$$

Take the inverse cosine ($\cos^{-1}$):
$$x = \cos^{-1}(0.63325)$$
$$x \approx 50.7^\circ$$

*(Self-Correction/Check using Law of Sines just to be sure: $\frac{\sin(45)}{6.396} = \frac{\sin(x)}{7}$. $\sin(x) = \frac{7\sin(45)}{6.396} = \frac{4.95}{6.396} = 0.7739$. $\arcsin(0.7739) = 50.7^\circ$. The calculation holds.)*

Final Answer:
Definition Blanks: sides, angles, right
When to Use Blanks: side, angle (for SAS); angle (for SSS)
Example 1: $x \approx 6.1$
Example 2: $x \approx 50.7^\circ$