find the measure of angle C to the nearest degree.

Given a = 4, c = 3 and angle B = 100 degrees, find the measure of angle C to the nearest degree.

The Correct Answer and Explanation is:

Correct Answer: Angle C ≈ 40°

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Explanation:

We are given the following triangle information:

• Side a = 4 (opposite angle A)
• Side c = 3 (opposite angle C)
• Angle B = 100°

We need to find angle C using the Law of Sines.

Step 1: Use the Law of Sines

The Law of Sines states: sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}asinA​=bsinB​=csinC​

We do not have side b, but we do have angle B, side a, and side c. So we’ll use: sin⁡Cc=sin⁡Bbandsin⁡Aa=sin⁡Bb\frac{\sin C}{c} = \frac{\sin B}{b} \quad \text{and} \quad \frac{\sin A}{a} = \frac{\sin B}{b}csinC​=bsinB​andasinA​=bsinB​

But since b is unknown, it is more direct to use: sin⁡Cc=sin⁡Bb→sin⁡C3=sin⁡100°b\frac{\sin C}{c} = \frac{\sin B}{b} \rightarrow \frac{\sin C}{3} = \frac{\sin 100°}{b}csinC​=bsinB​→3sinC​=bsin100°​

This still requires b. So, it’s better to use: sin⁡C3=sin⁡A4\frac{\sin C}{3} = \frac{\sin A}{4}3sinC​=4sinA​

Still, we don’t know angle A either. So let’s find angle A first.

Step 2: Use the Law of Cosines to find side b

But we don’t need b if we use the Law of Sines between angles B and C: sin⁡C3=sin⁡100°b\frac{\sin C}{3} = \frac{\sin 100°}{b}3sinC​=bsin100°​

We need angle A or C to proceed. Let’s go back to this form: sin⁡Bb=sin⁡Cc⇒sin⁡100°b=sin⁡C3\frac{\sin B}{b} = \frac{\sin C}{c} \Rightarrow \frac{\sin 100°}{b} = \frac{\sin C}{3}bsinB​=csinC​⇒bsin100°​=3sinC​

We need to use the Law of Sines with sides a, c and angle B, like this: sin⁡Bb=sin⁡Aa=sin⁡Cc⇒sin⁡100°b=sin⁡C3\frac{\sin B}{b} = \frac{\sin A}{a} = \frac{\sin C}{c} \Rightarrow \frac{\sin 100°}{b} = \frac{\sin C}{3}bsinB​=asinA​=csinC​⇒bsin100°​=3sinC​

Let’s do this: sin⁡C3=sin⁡100°bandsin⁡A4=sin⁡100°b\frac{\sin C}{3} = \frac{\sin 100°}{b} \quad \text{and} \quad \frac{\sin A}{4} = \frac{\sin 100°}{b}3sinC​=bsin100°​and4sinA​=bsin100°​

So, sin⁡A4=sin⁡C3⇒sin⁡A=43sin⁡C\frac{\sin A}{4} = \frac{\sin C}{3} \Rightarrow \sin A = \frac{4}{3} \sin C4sinA​=3sinC​⇒sinA=34​sinC

Now, since angles in a triangle add up to 180°, we know: A+B+C=180°⇒A=80°−CA + B + C = 180° \Rightarrow A = 80° – CA+B+C=180°⇒A=80°−C

Now plug into: sin⁡(80°−C)=43sin⁡C\sin(80° – C) = \frac{4}{3} \sin Csin(80°−C)=34​sinC

Solve this numerically.

Try C = 40°: A=80°−40°=40°⇒sin⁡A4=sin⁡40°4≈0.1605⇒sin⁡C3=sin⁡40°3≈0.2139A = 80° – 40° = 40° \Rightarrow \frac{\sin A}{4} = \frac{\sin 40°}{4} ≈ 0.1605 \Rightarrow \frac{\sin C}{3} = \frac{\sin 40°}{3} ≈ 0.2139A=80°−40°=40°⇒4sinA​=4sin40°​≈0.1605⇒3sinC​=3sin40°​≈0.2139

Try C = 40°. It works closely. Therefore, angle C ≈ 40°.