A triangle has sides A, B, and C. Sides A and B are of lengths 7 and 5, respectively, and the angle between A and B is (5pi)/8 . What is the length of side C?

Aug 2, 2017

$c = 10$

Explanation:

Well this is a very tricky one, i wish i can draw the Triangle to explain better, but with the theory i hope you will understand..

Firstly Note that the Angle is in Radians..

If you observe closely,

Angel between A and B is $\textcolor{w h i t e}{x} 5 \frac{\pi}{8}$

You're aware that there are only 3 angels which are $A , B , C$

Now again; What is the Angle between $A \mathmr{and} B$ ?

Simple, the answer is $C$

Cause it's a Triangle so the angle is C, since it's in between the two angles..

It's just a trick to push you off course from the desired Answer..

Hence we have as follows;

${C}^{o} = 5 \frac{\pi}{8}$

$a = 7$

$b = 5$

c = ?

Before we proceed further, remember first i said Note that the Angle is in Radians..

We must first convert from Radians to Degree..

Using this formula

$D e g r e e \Rightarrow r a \mathrm{di} a n s \times \frac{180}{\pi}$

Hence $\Rightarrow$ ${C}^{o} = \frac{5 \pi}{8} \times \frac{180}{\pi}$

${C}^{o} = \frac{5 \cancel{\pi}}{8} \times \frac{180}{\cancel{\pi}}$

${C}^{o} = \frac{5}{8} \times \frac{180}{1}$

${C}^{o} = \frac{5 \times 180}{8}$

${C}^{o} = \frac{900}{8}$

${C}^{o} = \frac{900}{8}$

${C}^{o} = 112.5$

${C}^{o} \approx 113$

Now to find c we shall using the Cosine Rule Formula

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b C o s \left(C\right)$

${c}^{2} = {7}^{2} + {5}^{2} - 2 \times 7 \times 5 C o s \left(113\right)$

${c}^{2} = 49 + 25 - 70 \left(- 0.39073\right)$

${c}^{2} = 74 + 27.3511$

${c}^{2} = 101.3511$

${c}^{2} \approx 101$

$c = \sqrt{101}$

$c = 10.04987$

$c \approx 10$

$\therefore c = 10 \to A n s w e r$