Solution:

Given:

• $A = 35°$
• $B = 65°$
• $c = 10$ cm
• Find: $a$

Step 1: Identify the case

This is AAS (two angles and one side), so use the Law of Sines.

Step 2: Find the third angle $C = 180° - A - B = 180° - 35° - 65° = 80°$

Step 3: Apply Law of Sines $\frac{a}{\sin A} = \frac{c}{\sin C}$

Substitute: $\frac{a}{\sin 35°} = \frac{10}{\sin 80°}$

Step 4: Solve for $a$ $a = \frac{10 \sin 35°}{\sin 80°}$

Calculate: $a = \frac{10 \times 0.5736}{0.9848} \approx \frac{5.736}{0.9848} \approx 5.82 \text{ cm}$

Answer: $a \approx 5.82$ cm

Verification:

• All angles sum to $180°$ ✓
• $a < c$ makes sense since $A < C$ (smaller angle opposite smaller side) ✓

Solution:

Part (a): Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$

$\frac{12}{\sin 35°} = \frac{15}{\sin B}$

$\sin B = \frac{15 \sin 35°}{12} = \frac{15(0.5736)}{12} = \frac{8.604}{12} = 0.717$

$B = \sin^{-1}(0.717) \approx 45.8°$

Note: There could be another solution $B' = 180° - 45.8° = 134.2°$, but we need to check if it's valid.

If $B = 134.2°$, then $A + B = 35° + 134.2° = 169.2° < 180°$, so this is also possible.

However, since $b > a$ and angle $B$ is opposite the larger side, we expect $B > A$.

Both solutions satisfy this, so we have the ambiguous case. Let's take $B \approx 45.8°$ as the acute solution.

Part (b): $C = 180° - A - B = 180° - 35° - 45.8° = 99.2°$

Part (c): Using Law of Sines:

$\frac{c}{\sin C} = \frac{a}{\sin A}$

$c = \frac{a \sin C}{\sin A} = \frac{12 \sin 99.2°}{\sin 35°} = \frac{12(0.9877)}{0.5736} \approx 20.7$

Solution:

Given:

• $a = 8$
• $b = 5$
• $C = 60°$
• Find: $c$

Step 1: Identify the case

This is SAS (two sides and the included angle), so use the Law of Cosines.

Step 2: Apply Law of Cosines $c^2 = a^2 + b^2 - 2ab\cos C$

Substitute: $c^2 = 8^2 + 5^2 - 2(8)(5)\cos 60°$

Step 3: Calculate

$c^2 = 64 + 25 - 80 \cdot \frac{1}{2}$ $c^2 = 64 + 25 - 40$ $c^2 = 49$ $c = 7$

Answer: $c = 7$

Bonus: Find angle $A$

Now use Law of Sines: $\frac{\sin A}{a} = \frac{\sin C}{c}$ $\frac{\sin A}{8} = \frac{\sin 60°}{7}$ $\sin A = \frac{8 \sin 60°}{7} = \frac{8 \times \frac{\sqrt{3}}{2}}{7} = \frac{4\sqrt{3}}{7} \approx 0.9897$ $A = \arcsin(0.9897) \approx 81.8°$

Verification:

• $c^2 = 49$ gives $c = 7$ ✓
• When $C = 60°$, $\cos 60° = 0.5$ ✓

Solution:

Part (a): Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$

$14^2 = 8^2 + 11^2 - 2(8)(11)\cos C$

$196 = 64 + 121 - 176\cos C$

$196 = 185 - 176\cos C$

$11 = -176\cos C$

$\cos C = -\frac{11}{176} = -0.0625$

$C = \cos^{-1}(-0.0625) \approx 93.6°$

Part (b): Area formula using two sides and included angle:

First, we need to find one of the other angles. Let's find angle $A$:

$a^2 = b^2 + c^2 - 2bc\cos A$

$64 = 121 + 196 - 2(11)(14)\cos A$

$64 = 317 - 308\cos A$

$-253 = -308\cos A$

$\cos A = \frac{253}{308} = 0.821$

$A = \cos^{-1}(0.821) \approx 34.7°$

Now use: Area $= \frac{1}{2}bc\sin A$

$= \frac{1}{2}(11)(14)\sin 34.7°$

$= \frac{1}{2}(154)(0.569)$

$\approx 43.8$ square units

Solution:

Given:

• $a = 20$
• $b = 15$
• $c = 25$
• Find: All angles

Step 1: Identify the case

This is SSS (all three sides), so use the Law of Cosines.

Step 2: Find angle $C$ (largest angle, opposite longest side)

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Substitute: $\cos C = \frac{20^2 + 15^2 - 25^2}{2(20)(15)}$ $\cos C = \frac{400 + 225 - 625}{600}$ $\cos C = \frac{0}{600} = 0$

Therefore: $C = \arccos(0) = 90°$

This is a right triangle!

Step 3: Find angle $A$

Now we can use Law of Sines: $\frac{\sin A}{a} = \frac{\sin C}{c}$ $\frac{\sin A}{20} = \frac{\sin 90°}{25}$ $\frac{\sin A}{20} = \frac{1}{25}$ $\sin A = \frac{20}{25} = \frac{4}{5} = 0.8$ $A = \arcsin(0.8) \approx 53.1°$

Step 4: Find angle $B$

$B = 180° - A - C = 180° - 53.1° - 90° = 36.9°$

Answer:

• $A \approx 53.1°$
• $B \approx 36.9°$
• $C = 90°$

Verification:

• $A + B + C \approx 53.1° + 36.9° + 90° = 180°$ ✓
• Check Pythagorean theorem: $20^2 + 15^2 = 400 + 225 = 625 = 25^2$ ✓
• This confirms it's a right triangle! ✓

Note: This is a 3-4-5 right triangle scaled by 5 (sides 15, 20, 25).