Question

question 15 (5 points)
find the area of the triangle with ( a = 12.9 ), ( b = 12.4 ), and ( c = 17.1 ). round to the nearest tenth.
( 80.1 ) units( ^{2} )
( 79.1 ) units( ^{2} )
( 79.7 ) units( ^{2} )
( 82.7 ) units( ^{2} )
question 16 (5 points)
find the exact value of ( cos \frac{11 pi}{6} ).
( \frac{1}{2} )
( -\frac{sqrt{3}}{2} )
( \frac{sqrt{3}}{2} )
undefined

Explanation:

Question 15

Step1: Calculate the semi - perimeter \(s\)

The formula for the semi - perimeter of a triangle with sides \(a\), \(b\), and \(c\) is \(s=\frac{a + b + c}{2}\).
Substitute \(a = 12.9\), \(b = 12.4\), and \(c = 17.1\) into the formula:
\(s=\frac{12.9+12.4 + 17.1}{2}=\frac{42.4}{2}=21.2\)

Step2: Use Heron's formula \(A=\sqrt{s(s - a)(s - b)(s - c)}\)

Substitute \(s = 21.2\), \(a = 12.9\), \(b = 12.4\), and \(c = 17.1\) into Heron's formula:
\(A=\sqrt{21.2(21.2-12.9)(21.2 - 12.4)(21.2-17.1)}\)
\(=\sqrt{21.2\times8.3\times8.8\times4.1}\)
\(=\sqrt{21.2\times(8.3\times8.8\times4.1)}\)
\(=\sqrt{21.2\times(73.04\times4.1)}\)
\(=\sqrt{21.2\times299.464}\)
\(=\sqrt{6348.6368}\approx79.7\)

We know that the cosine function has a period of \(2\pi\), and the formula \(\cos(x + 2k\pi)=\cos x\), \(k\in\mathbb{Z}\). Also, \(\cos(\theta)=\cos(2\pi-\theta)\)

We can write \(\frac{11\pi}{6}=2\pi-\frac{\pi}{6}\)

By the identity \(\cos(A - B)=\cos A\cos B+\sin A\sin B\) (where \(A = 2\pi\), \(B=\frac{\pi}{6}\)), and since \(\cos(2\pi)=1\), \(\sin(2\pi)=0\)

\(\cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos2\pi\cos\frac{\pi}{6}+\sin2\pi\sin\frac{\pi}{6}\)

\(=1\times\frac{\sqrt{3}}{2}+0\times\frac{1}{2}=\frac{\sqrt{3}}{2}\)

Answer:

\(79.7\ units^{2}\)

Question 16