Question

1. find the missing parts of the triangle abc if ( a = 60^circ ), ( b = 25 ) inches, and ( c = 32 ) inches. show all the details in a logical order.

Explanation:

Step1: Use the Law of Cosines to find side \( a \)

The Law of Cosines states that \( a^2 = b^2 + c^2 - 2bc\cos A \). Given \( A = 60^\circ \), \( b = 25 \) inches, \( c = 32 \) inches. First, calculate \( \cos 60^\circ=\frac{1}{2} \). Then substitute the values:

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So, \( a=\sqrt{849}\approx29.14 \) inches.

Step2: Use the Law of Sines to find angle \( B \)

The Law of Sines is \( \frac{\sin B}{b}=\frac{\sin A}{a} \). We know \( b = 25 \), \( A = 60^\circ \), \( a\approx29.14 \), and \( \sin 60^\circ=\frac{\sqrt{3}}{2}\approx0.8660 \). Rearranging for \( \sin B \):

$$ \sin B=\frac{b\sin A}{a}=\frac{25\times\sin 60^\circ}{29.14}=\frac{25\times0.8660}{29.14}\approx\frac{21.65}{29.14}\approx0.743 $$

Then, \( B=\arcsin(0.743)\approx47.9^\circ \approx 48^\circ \).

Step3: Find angle \( C \)

Since the sum of angles in a triangle is \( 180^\circ \), \( C = 180^\circ - A - B \). Substitute \( A = 60^\circ \) and \( B\approx48^\circ \):

$$ C = 180^\circ - 60^\circ - 48^\circ = 72^\circ $$

(We can also verify using Law of Sines for angle \( C \) to check consistency: \( \frac{\sin C}{c}=\frac{\sin A}{a} \), \( \sin C=\frac{32\times\sin 60^\circ}{29.14}\approx\frac{32\times0.8660}{29.14}\approx\frac{27.712}{29.14}\approx0.951 \), so \( C=\arcsin(0.951)\approx72^\circ \), which matches.)

Answer:

• Side \( a\approx29.14 \) inches (or \( \sqrt{849} \) inches)
• Angle \( B\approx48^\circ \)
• Angle \( C = 72^\circ \)