Question

find bc. round to the nearest tenth.
triangle with vertices b, a, c; side ab = 6, side ac = 13, angle at a is 91°
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Explanation:

Step1: Identify the Law of Cosines

We have a triangle with two sides and the included angle. The Law of Cosines formula for side \( a \) (opposite angle \( A \)) is \( a^2 = b^2 + c^2 - 2bc \cos A \). Here, we want to find \( BC \) (let's call it \( a \)), with \( b = 13 \), \( c = 6 \), and \( A = 91^\circ \).

Step2: Substitute the values into the formula

Substitute \( b = 13 \), \( c = 6 \), and \( A = 91^\circ \) into the formula:
\( BC^2 = 13^2 + 6^2 - 2 \times 13 \times 6 \times \cos(91^\circ) \)
First, calculate \( 13^2 = 169 \), \( 6^2 = 36 \), and \( 2 \times 13 \times 6 = 156 \).
\( \cos(91^\circ) \approx -0.01745 \) (using a calculator)
So, \( BC^2 = 169 + 36 - 156 \times (-0.01745) \)
\( BC^2 = 205 + 156 \times 0.01745 \)
\( 156 \times 0.01745 \approx 2.7222 \)
\( BC^2 \approx 205 + 2.7222 = 207.7222 \)

Step3: Take the square root and round

Take the square root of \( 207.7222 \): \( BC \approx \sqrt{207.7222} \approx 14.4 \) (rounded to the nearest tenth)

Answer:

\( 14.4 \)