Law of Cosine to Figure Area of a Triangle - Free Printable

Here are the step-by-step solutions for each problem on the worksheet. We will use the Law of Cosines to find the missing side length in each triangle.

The formula is:
$c^2 = a^2 + b^2 - 2ab \cos(C)$
*(Where $C$ is the angle between sides $a$ and $b$, and $c$ is the side opposite that angle.)*

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1.) Find side AB (let's call it $c$)

* Given: Side $AC = 6$, Side $BC = 4$, Angle $C = 22^\circ$.
* Step 1: Plug values into the formula:
$c^2 = 6^2 + 4^2 - 2(6)(4) \cos(22^\circ)$
* Step 2: Calculate squares and multiplication:
$c^2 = 36 + 16 - 48 \cos(22^\circ)$
* Step 3: Find $\cos(22^\circ) \approx 0.9272$:
$c^2 = 52 - 48(0.9272)$
$c^2 = 52 - 44.5056$
$c^2 = 7.4944$
* Step 4: Take the square root:
$c = \sqrt{7.4944} \approx 2.737$
* Rounding: To the nearest tenth, $c = 2.7$.

2.) Find side AC (let's call it $b$)

* Given: Side $AB = c$ (unknown), Side $BC = a = 5$, Side $AC = b$ (unknown). Wait, looking at the diagram:
* Side $BC = 5$
* Angle $B = 79^\circ$
* Angle $A = 43^\circ$
* We need to find side $AC$ (opposite angle B? No, side AC is opposite angle B? Let's check labels. Vertices are A, B, C. Side opposite B is AC. Side opposite A is BC. Side opposite C is AB.
* Actually, let's look closer. We have Side $BC=5$ (adjacent to angle B and C). We have Angle $B=79^\circ$ and Angle $A=43^\circ$.
* First, find Angle $C$: $180^\circ - 79^\circ - 43^\circ = 58^\circ$.
* Now we can use Law of Sines or Law of Cosines. The worksheet says "Law of Cosines", but Law of Sines is easier here. However, if we must use Cosines, we need two sides and an included angle. We only have one side ($BC=5$).
* Let's re-read the diagram carefully.
* Triangle with vertices A, B, C.
* Side $BC$ (left side) is labeled 5.
* Angle at B is $79^\circ$.
* Angle at A is $43^\circ$.
* We need to find side $AC$ (side $b$) or side $AB$ (side $c$)? Usually, you solve for the side opposite the known angle if you have another side.
* Let's assume the question asks for side $AC$ (opposite angle B) or side $AB$ (opposite angle C).
* Actually, usually these problems give SAS (Side-Angle-Side). This looks like AAS (Angle-Angle-Side).
* Let's calculate Angle $C = 180 - 79 - 43 = 58^\circ$.
* Using Law of Sines is standard for AAS: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.
* $\frac{\sin(43)}{5} = \frac{\sin(79)}{b}$ (finding side AC, opposite B).
* $b = \frac{5 \cdot \sin(79)}{\sin(43)} = \frac{5 \cdot 0.9816}{0.6820} \approx 7.19$.
* Let's check if there is a different interpretation. Maybe the side labeled 5 is $AC$? No, it's next to vertex B and C. It's side $a$ (opposite A)? No, standard notation: side $a$ is opposite A. Side $BC$ is opposite A. So $a=5$.
* If $a=5$ (opposite A=43), and we want side $c$ (AB, opposite C=58):
$c = \frac{5 \cdot \sin(58)}{\sin(43)} = \frac{5 \cdot 0.8480}{0.6820} \approx 6.22$.
* If we want side $b$ (AC, opposite B=79):
$b = \frac{5 \cdot \sin(79)}{\sin(43)} \approx 7.2$.
* Looking at the visual length, side AC (opposite 79) should be the longest. Side AB (opposite 58) is middle. Side BC (opposite 43) is shortest (5).
* The prompt asks to use Law of Cosines. Is it possible the label "5" refers to side $AB$? If side $c=5$, angle $B=79$, angle $A=43$, then angle $C=58$. Then we have ASA. Still requires finding a side first.
* Let's assume the standard case where we might need to find the third angle first. Since the worksheet title is "Law of Cosines", but problem 2 is AAS/ASA, it's slightly tricky. However, often "Law of Cosines Worksheets" include mixed problems. Let's provide the answer for the side opposite the largest angle (AC) as it's the most common "solve for x" variable, or perhaps side AB. Let's look at the position of the "5". It is on the left leg. That is side $BC$ (or $a$). The angle at the bottom left is B ($79^\circ$). The angle at the bottom right is A ($43^\circ$).
* Let's solve for side $AC$ (side $b$) and side $AB$ (side $c$). Usually, the unknown side is the one not touching the known angles if it's SAS, but here it's AAS.
* Let's assume the question implies finding the side opposite the given angle B? Or maybe the side connecting the two known angles? No, that's side AB.
* Let's calculate side $AC$ (opposite $79^\circ$): $\approx 7.2$.
* Let's calculate side $AB$ (opposite $58^\circ$): $\approx 6.2$.
* Without a specific variable marked "x", I will provide the length of side AC as it is the side opposite the explicitly labeled angle $79^\circ$ which is adjacent to the known side. Wait, no. Side 5 is adjacent to 79. Side AC is opposite 79.
* Actually, let's look at Problem 1. It asked for the side opposite the angle. Problem 3 asks for side opposite angle. Problem 2 likely asks for side $AC$ (opposite B) or $AB$ (opposite C). Let's provide AC as the primary answer, but note that AB is also solvable. Given the layout, usually, you solve for the side opposite the "top" angle or the remaining side. Let's go with AC ≈ 7.2.

*(Self-Correction: In many textbooks, if only one side is given with two angles, you use Law of Sines. If the instruction strictly says Law of Cosines, you might have to derive it, but Sines is the intended path for AAS. I will provide the value for side AC.)*

3.) Find side BC (let's call it $a$)

* Given: Side $AB = 7$, Side $AC = 10$, Angle $A = 68^\circ$.
* Step 1: Formula: $a^2 = 7^2 + 10^2 - 2(7)(10) \cos(68^\circ)$
* Step 2: Calculate:
$a^2 = 49 + 100 - 140 \cos(68^\circ)$
* Step 3: $\cos(68^\circ) \approx 0.3746$:
$a^2 = 149 - 140(0.3746)$
$a^2 = 149 - 52.444$
$a^2 = 96.556$
* Step 4: Square root:
$a = \sqrt{96.556} \approx 9.826$
* Rounding: To the nearest tenth, $a = 9.8$.

4.) Find side AC (let's call it $b$)

* Given: Side $AB = 6$, Side $BC = 6$, Side $AC$ is unknown? Wait.
* Side $AB = 6$ (left).
* Side $BC = 6$ (bottom).
* Side $AC = 5$ (right).
* This is an SSS (Side-Side-Side) triangle. There is no angle given.
* Usually, Law of Cosines worksheets ask to find an angle when three sides are given.
* Let's check the text: "Answer to the nearest tenth." Angles are usually degrees, sides are lengths. But "nearest tenth" applies to both.
* If the task is to find an angle, which one? Usually the largest angle (opposite the longest side) or a specific one. The longest side is 6 (tie). The shortest is 5.
* Let's assume we need to find Angle B (between the two sides of length 6)? Or Angle A? Or Angle C?
* Let's look at the other problems. They all find a side. Problem 4 has all 3 sides labeled: 6, 6, 5. It does not have a question mark.
* Perhaps I am misreading the image. Let me look really closely at Crop 4.
* Side $AB = 6$. Side $BC = 6$. Side $AC = 5$.
* There is no "x" or "?".
* However, in similar worksheets, if 3 sides are given, the task is often to find the largest angle or all angles.
* Let's calculate Angle $B$ (opposite side 5):
$5^2 = 6^2 + 6^2 - 2(6)(6) \cos(B)$
$25 = 36 + 36 - 72 \cos(B)$
$25 = 72 - 72 \cos(B)$
$-47 = -72 \cos(B)$
$\cos(B) = 47/72 \approx 0.6528$
$B = \arccos(0.6528) \approx 49.2^\circ$.
* Let's calculate Angle $A$ (opposite side 6):
$6^2 = 6^2 + 5^2 - 2(6)(5) \cos(A)$
$36 = 36 + 25 - 60 \cos(A)$
$0 = 25 - 60 \cos(A)$
$\cos(A) = 25/60 \approx 0.4167$
$A = \arccos(0.4167) \approx 65.4^\circ$.
* Angle $C$ is the same as $A$ because it's isosceles ($65.4^\circ$).
* Check sum: $49.2 + 65.4 + 65.4 = 180$. Correct.
* Since the worksheet title is "Law of Cosines" and previous questions found sides, this one likely asks for an angle. Without a specific target, finding the vertex angle B is the most distinct calculation. Or perhaps the base angles. I will provide Angle B = 49.2° as the primary answer, but note the others. *Wait*, looking at the pattern, maybe one of the numbers is actually the unknown? No, they are clearly printed. I will assume the question asks to find Angle B.

5.) Find side AB (let's call it $c$)

* Given: Side $BC = 7$, Side $AC = 7$, Angle $C = 130^\circ$.
* Step 1: Formula: $c^2 = 7^2 + 7^2 - 2(7)(7) \cos(130^\circ)$
* Step 2: Calculate:
$c^2 = 49 + 49 - 98 \cos(130^\circ)$
* Step 3: $\cos(130^\circ) \approx -0.6428$:
$c^2 = 98 - 98(-0.6428)$
$c^2 = 98 + 62.994$
$c^2 = 160.994$
* Step 4: Square root:
$c = \sqrt{160.994} \approx 12.688$
* Rounding: To the nearest tenth, $c = 12.7$.

6.) Find side AC (let's call it $b$)

* Given: Side $AB = 9.6$, Side $BC = 8$, Angle $B = 102^\circ$.
* Step 1: Formula: $b^2 = 9.6^2 + 8^2 - 2(9.6)(8) \cos(102^\circ)$
* Step 2: Calculate squares:
$9.6^2 = 92.16$
$8^2 = 64$
* Step 3: Multiply terms:
$2(9.6)(8) = 153.6$
* Step 4: $\cos(102^\circ) \approx -0.2079$:
$b^2 = 92.16 + 64 - 153.6(-0.2079)$
$b^2 = 156.16 + 31.933$
$b^2 = 188.093$
* Step 5: Square root:
$b = \sqrt{188.093} \approx 13.714$
* Rounding: To the nearest tenth, $b = 13.7$.

7.) Find side AB (let's call it $c$)

* Given: Side $BC = 4$, Side $AC = 10$, Angle $C = 108^\circ$.
* Step 1: Formula: $c^2 = 4^2 + 10^2 - 2(4)(10) \cos(108^\circ)$
* Step 2: Calculate:
$c^2 = 16 + 100 - 80 \cos(108^\circ)$
* Step 3: $\cos(108^\circ) \approx -0.3090$:
$c^2 = 116 - 80(-0.3090)$
$c^2 = 116 + 24.72$
$c^2 = 140.72$
* Step 4: Square root:
$c = \sqrt{140.72} \approx 11.862$
* Rounding: To the nearest tenth, $c = 11.9$.

8.) Find side AC (let's call it $b$)

* Given: Side $AB = 8$, Side $BC = 5$, Angle $B = 130^\circ$.
* Step 1: Formula: $b^2 = 8^2 + 5^2 - 2(8)(5) \cos(130^\circ)$
* Step 2: Calculate:
$b^2 = 64 + 25 - 80 \cos(130^\circ)$
* Step 3: $\cos(130^\circ) \approx -0.6428$:
$b^2 = 89 - 80(-0.6428)$
$b^2 = 89 + 51.424$
$b^2 = 140.424$
* Step 4: Square root:
$b = \sqrt{140.424} \approx 11.850$
* Rounding: To the nearest tenth, $b = 11.9$.

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Final Answer:
1.) 2.7
2.) 7.2 (Side AC) or 6.2 (Side AB) — *Note: Problem 2 requires Law of Sines as it is AAS. Side AC is opposite the 79° angle.*
3.) 9.8
4.) 49.2° (Angle B) — *Note: Problem 4 provides 3 sides (SSS), so the unknown is an angle. Angle B is opposite the side of length 5.*
5.) 12.7
6.) 13.7
7.) 11.9
8.) 11.9