Question

question 5 of 26
given the triangle below, what is ( mangle c ), rounded to the nearest tenth?
triangle with vertices a, b, c: a and c on base (ac = 12), b at top; ab = 8, bc = 10; note: triangle not drawn to scale
options:
a. ( 38.6^circ )
b. ( 48.4^circ )
c. ( 46.2^circ )
d. ( 41.4^circ )

Explanation:

Step1: Identify the Law to Use

We have a triangle with sides \( a = 8 \), \( b = 10 \), \( c = 12 \) (opposite angles \( A \), \( B \), \( C \) respectively). To find \( m\angle C \), we can use the Law of Cosines: \( \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab} \).

Step2: Substitute the Values

Substitute \( a = 8 \), \( b = 10 \), \( c = 12 \) into the formula:
\( \cos C=\frac{8^{2}+10^{2}-12^{2}}{2\times8\times10}=\frac{64 + 100-144}{160}=\frac{20}{160}=\frac{1}{8} = 0.125 \)

Step3: Find the Angle

Take the inverse cosine: \( C=\cos^{-1}(0.125)\approx82.8^{\circ}\)? Wait, no, wait. Wait, maybe I mixed up the sides. Wait, in triangle \( ABC \), side \( AB = 8 \), \( BC = 10 \), \( AC = 12 \). So angle \( C \) is between \( AC = 12 \) and \( BC = 10 \), so the sides adjacent to angle \( C \) are \( 10 \) and \( 12 \), and the side opposite is \( 8 \). So Law of Cosines: \( AB^{2}=AC^{2}+BC^{2}-2\times AC\times BC\times\cos C \). So \( 8^{2}=12^{2}+10^{2}-2\times12\times10\times\cos C \). Then \( 64 = 144 + 100-240\cos C \). Then \( 64=244 - 240\cos C \). Then \( - 180=-240\cos C \). Then \( \cos C=\frac{180}{240}=0.75 \). Then \( C=\cos^{-1}(0.75)\approx41.4^{\circ} \).

Answer:

D. \( 41.4^{\circ} \)