Question

question 13 (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) 7.3 m
b) 14.5 m
c) 9.9 m
d) 11.2 m

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:
\(c^{2}=(6.9)^{2}+(15.7)^{2}-2\times6.9\times15.7\times\cos(39^{\circ})\)

Step2: Calculate each term

First, \((6.9)^{2}=47.61\), \((15.7)^{2}=246.49\).
\(\cos(39^{\circ})\approx0.7771\), and \(2\times6.9\times15.7 = 216.06\).
Then \(216.06\times0.7771\approx168.9\)

Step3: Compute \(c^{2}\)

\(c^{2}=47.61 + 246.49-168.9\)
\(c^{2}=309.58\)

Step4: Find \(c\)

\(c=\sqrt{309.58}\approx17.6\) (Wait, no, let's re - calculate.
Wait, correct calculation:
\(c^{2}=6.9^{2}+15.7^{2}-2\times6.9\times15.7\times\cos(39^{\circ})\)
\(6.9^{2}=47.61\), \(15.7^{2}=246.49\)
\(2\times6.9\times15.7 = 216.06\), \(\cos(39^{\circ})\approx0.7771\)
\(216.06\times0.7771 = 216.06\times(0.7 + 0.07+0.0071)=216.06\times0.7+216.06\times0.07 + 216.06\times0.0071=151.242+15.1242 + 1.534=167.9\)
\(c^{2}=47.61+246.49 - 167.9=126.2\)
\(c=\sqrt{126.2}\approx11.2\)

Answer:

d) \(11.2\ m\)