Question

solve the triangle. round the lengths of sides to the nearest tenth and angles to the nearest degree. c≈□ (type an integer or decimal rounded to the nearest tenth as needed.) a≈□° (round to the nearest degree as needed.) b≈□° (round to the nearest degree as needed.)

Explanation:

Step1: Find side \( c \) using the Law of Cosines

The Law of Cosines states that for a triangle with sides \( a \), \( b \), \( c \) and opposite angles \( A \), \( B \), \( C \) respectively, \( c^{2}=a^{2}+b^{2}-2ab\cos C \). Here, \( a = 5 \), \( b = 3 \), and \( C=97^{\circ} \). First, calculate \( \cos(97^{\circ}) \). Using a calculator, \( \cos(97^{\circ})\approx - 0.1219 \). Then:

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Taking the square root of both sides, \( c=\sqrt{37.657}\approx6.1 \)

Step2: Find angle \( A \) using the Law of Sines

The Law of Sines states that \( \frac{a}{\sin A}=\frac{c}{\sin C} \). We know \( a = 5 \), \( c\approx6.1 \), and \( C = 97^{\circ} \), \( \sin(97^{\circ})\approx0.9925 \). So:

$$ \sin A=\frac{a\sin C}{c}=\frac{5\times0.9925}{6.1}=\frac{4.9625}{6.1}\approx0.8135 $$

Then \( A=\sin^{-1}(0.8135)\approx54^{\circ} \) (since \( A \) is an acute angle in the triangle, we take the inverse sine of \( 0.8135 \))

Step3: Find angle \( B \) using the angle sum property of a triangle

The sum of angles in a triangle is \( 180^{\circ} \). So \( B=180^{\circ}-A - C \). We know \( A\approx54^{\circ} \) and \( C = 97^{\circ} \), so:

$$ B=180 - 54-97=29^{\circ} $$

Answer:

\( c\approx6.1 \), \( A\approx54^{\circ} \), \( B\approx29^{\circ} \)