Question

Two sides of a triangle have measures 3 ft and 6 ft. Also, these sides form a vertex whose angle measures 60 degrees. Calculate the missing attributes of the triangle.

Answer 1

Answer:

A. The length of the third side is approximately 5 feet.

B. A = $88^{o}$, B = $60^{o}$ and C = $32^{o}$.

Step-by-step explanation:

Let the triangle be ABC. Given two sides and an included angle, let us apply the cosine rule to determine the third length.

A. Let side a = 6 feet and c = 3 feet, thus;

$b^{2}$ = $a^{2}$ + $c^{2}$ - 2ac Cos B

  = $6^{2}$ + $3^{2}$ - 2(6 x 3) Cos $60^{o}$

  = 36 + 9 - 36 x 0.5

  = 45 - 18

$b^{2}$ = 27

b = $\sqrt{27}$

  = 5.1962

b  = 5.2 feet

b ≅ 5 feet

The length of the third side is approximately 5 feet.

B. Given that B = $60^{o}$, then let us apply the Sine rule to determined the measure of A.

$\frac{a}{SinA}$ = $\frac{b}{SinB}$

So that,

$\frac{6}{Sin A}$ = $\frac{5.2}{Sin60^{o} }$

Sin A = $\frac{0.886*6}{5.2}$

Sin A = 0.99923

A = $Sin^{-1}$ 0.99923

  = 87.75

A ≅ $88^{o}$

Since the sum of angle in a triangle = $180^{o}$.

Then,

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

$88^{o}$ + $60^{o}$ + C = $180^{o}$

$148^{o}$ + C = $180^{o}$

C = $180^{o}$ - $148^{o}$

     = $32^{o}$

C = $32^{o}$

Thus,

A = $88^{o}$, B = $60^{o}$ and C = $32^{o}$.