Question

question 6 (mandatory) (1 point) saved
in \\( \triangle a b c, a = 6.9 \mathrm { m }, b = 15.7 \mathrm { m } \\), and \\( \angle c = 39 ^ { \circ } \\). determine \\( c \\) to the nearest tenth of a
metre.
\\( \bigcirc \\) a) \\( 11.2 \mathrm { m } \\)
\\( \bigcirc \\) b) \\( 9.9 \mathrm { m } \\)
\\( \bigcirc \\) c) \\( 14.5 \mathrm { m } \\)
\\( \bigcirc \\) d) \\( 7.3 \mathrm { m } \\)
question 7 (mandatory) (1 point) saved
determine the length of \\( x \\), to the nearest tenth of a kilometre.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\).
Substitute \(a = 6.9\), \(b = 15.7\), and \(C = 39^{\circ}\) into the formula:
\(c^{2}=6.9^{2}+15.7^{2}-2\times6.9\times15.7\times\cos(39^{\circ})\)
First, calculate \(6.9^{2}=47.61\), \(15.7^{2}=246.49\), and \(\cos(39^{\circ})\approx0.7771\).
Then \(2\times6.9\times15.7 = 2\times108.33=216.66\).
\(c^{2}=47.61 + 246.49-216.66\times0.7771\)
\(c^{2}=294.1-216.66\times0.7771\)
Calculate \(216.66\times0.7771\approx168.4\)
\(c^{2}=294.1 - 168.4=125.7\)

Step2: Solve for \(c\)

Take the square - root of both sides: \(c=\sqrt{125.7}\approx11.2\)

Answer:

A. \(11.2\ m\)