Question

solve the triangle. round the lengths of sides to the nearest tenth and angles to the nearest degree. triangle image with angle c = 94°, side a = 5, side b = 3 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: Use the Law of Cosines to find side \( c \)

The Law of Cosines states that \( c^2 = a^2 + b^2 - 2ab\cos(C) \). Given \( a = 5 \), \( b = 3 \), and \( C = 94^\circ \), we first calculate \( \cos(94^\circ) \approx -0.0698 \). Then:

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Taking the square root, \( c \approx \sqrt{36.094} \approx 6.0 \)

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

The Law of Sines is \( \frac{\sin(A)}{a} = \frac{\sin(C)}{c} \). So \( \sin(A) = \frac{a\sin(C)}{c} \). Substituting \( a = 5 \), \( C = 94^\circ \), \( c \approx 6.0 \):

$$ \sin(A)=\frac{5\times\sin(94^\circ)}{6.0}\approx\frac{5\times0.9976}{6.0}\approx\frac{4.988}{6.0}\approx0.8313 $$

Then \( A \approx \arcsin(0.8313) \approx 56^\circ \)

Step3: Find angle \( B \)

Since the sum of angles in a triangle is \( 180^\circ \), \( B = 180^\circ - A - C \). Substituting \( A \approx 56^\circ \), \( C = 94^\circ \):

$$ B = 180 - 56 - 94 = 30^\circ $$

Answer:

\( c \approx 6.0 \)
\( A \approx 56^\circ \)
\( B \approx 30^\circ \)