Question

A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes) between each of the sides?

Answer 1

A triangle is an example of a class of figures referred to as plane shapes. It has three straight sides and three internal angles which sum up to $180^{o}$. The measures of the internal angles of the triangle given in the question are A = $52.6^{o}$, B = $59.4^{o}$, and C = $68^{o}$.

A triangle is an example of a class of figures referred to as plane shapes. It has three straight sides and three internal angles which sum up to $180^{o}$.

Considering the given question, let the sides of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the Cosine rule to have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

So that;

$7^{2}$ = $6^{2}$ + $(6.5)^{2}$ - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

               = 29.25

Cos C = $\frac{29.25}{78}$

         = 0.375

C = $Cos^{-1}$ 0.375

   = 67.9757

C = $68^{o}$

Apply the Sine rule to determine the value of B,

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{6.5}{Sin B}$ = $\frac{7}{Sin 68}$

SIn B = $\frac{6.5 *Sin 68}{7}$

         = 0.861

B = $Sin^{-1}$ 0.861

   = 59.43

B = $59.4^{o}$

Thus to determine the value of A, we have;

A + B + C = $180^{o}$

A + $59.4^{o}$ + $68^{o}$ = $180^{o}$

A = $180^{o}$ - 127.4

  = 52.6

A = $52.6^{o}$

Therefore the sizes of the internal angles of the triangle are: A = $52.6^{o}$, B = $59.4^{o}$, and C = $68^{o}$.

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

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