Question

question 6 (mandatory) (1 point)
in △abc, a = 6.9 m, b = 15.7 m, and ∠c = 39°. determine c to the nearest tenth of a metre.

a) 11.2 m
b) 9.9 m
c) 14.5 m
d) 7.3 m

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\).
Given \(a = 6.9\), \(b = 15.7\), and \(C=39^{\circ}\).
Substitute the values 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 \(2\times6.9\times15.7 = 217.86\).
\(c^{2}=47.61 + 246.49-217.86\times0.7771\)

Step2: Calculate the value of \(c^{2}\)

\(47.61+246.49 = 294.1\)
\(217.86\times0.7771\approx169.3\)
\(c^{2}=294.1-169.3=124.8\)

Step3: Find the value of \(c\)

Take the square - root of \(c^{2}\), \(c=\sqrt{124.8}\approx11.2\)

Answer:

A. \(11.2\ m\)