To solve triangle $ABC$, where:

• $b = 60$
• $c = 38$
• $\angle A = 78^\circ$

We can apply the Law of Cosines and Law of Sines.

Step 1: Use the Law of Cosines to find $a$

The Law of Cosines states that:

$a^2 = b^2 + c^2 - 2bc \cdot \cos(\angle A)$

Substitute the known values:

$a^2 = 60^2 + 38^2 - 2 \cdot 60 \cdot 38 \cdot \cos(78^\circ)$

First, calculate the cosine of 78°:

$\cos(78^\circ) \approx 0.2079$

Now substitute this value:

$a^2 = 3600 + 1444 - 2 \cdot 60 \cdot 38 \cdot 0.2079$

$a^2 = 5044 - 1493.416$

$a^2 \approx 3549.584$

Now, take the square root of both sides:

$a \approx \sqrt{3549.584} \approx 59.6$

So, $a \approx 59.6$.

Step 2: Use the Law of Sines to find $\angle B$

The Law of Sines states:

$\frac{a}{\sin(\angle A)} = \frac{b}{\sin(\angle B)}$

We already know $a$, $b$, and $\angle A$, so we can solve for $\angle B$:

$\frac{59.6}{\sin(78^\circ)} = \frac{60}{\sin(\angle B)}$

First, calculate $\sin(78^\circ)$:

$\sin(78^\circ) \approx 0.9781$

Now substitute:

$\frac{59.6}{0.9781} = \frac{60}{\sin(\angle B)}$

$60.9 \approx \frac{60}{\sin(\angle B)}$

Now solve for $\sin(\angle B)$:

$\sin(\angle B) \approx \frac{60}{60.9} \approx 0.9852$

So, $\angle B \approx \sin^{-1}(0.9852) \approx 81.7^\circ$.

Step 3: Find $\angle C$

Since the sum of the angles in a triangle is 180°, we can find $\angle C$:

$\angle C = 180^\circ - \angle A - \angle B$

$\angle C = 180^\circ - 78^\circ - 81.7^\circ$

$\angle C \approx 20.3^\circ$

Final Answers:

• $a \approx 59.6$
• $\angle B \approx 81.7^\circ$
• $\angle C \approx 20.3^\circ$

Let me know if you need further details!

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Related Questions:

1. How do the Law of Cosines and Law of Sines differ in use?
2. What are the conditions for applying the Law of Cosines in a triangle?
3. What would happen if the triangle had an obtuse angle?
4. Can the Law of Sines always be used to find all angles?
5. How can you verify the angles in a triangle?

Tip: Remember to always check that the angles in a triangle add up to 180°, as it’s a key property of all triangles!