precalc hw

Question

Let the angles of a triangle be α, β, γ, with opposite sides of length a, b, and c, respectively. Use the Law of Cosines to find the remaining side and one of the other angles. (Round your answers to one decimal place.)
γ = 115°; a = 15; b = 18

Answer 1

The Law of Cosines states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus their double product multiplied by the cosine of the angle they form and as you will see, the Pythagorean Theorem for right triangles is a only a particular case of this more general theorem.

 

In the triangle abc you presented, this law is expressed:  c^2=a^2 + b^2 -2ab(cos y)

Substituting your values on the above equation:              c^2= 15^2 + 18^2 -2(15x18)(cos 115)

                                                                                   c^2= 549 - 540(-0.4226)

                                                                                   c^2= 777.20

                                                                                       c= 27.88

 

In the case of y=90 degrees, a right triangle the Law of Cosines becomes the well known Pythagorean Theorem because cos 90=0, thus:    c^2= a^2 + b^2

 

Best regards!

 

 

 

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Answer 2

α is opposite to side a.

β is opposite to side b.

γ is opposite to side c.

 

We have one known angle and 2 known sides.  We can use law of cosine to find one of the missing angles and the last side of the triangle.

 

First, we can find side c.

 

c² = a² + b² - 2abCos(γ)

 

 

Plug in the known values.  Once you have c, use the law of cosine once more to find β.

 

b² = a² + c² - 2acCos(β)

 

b² - a² - c² = -2acCos(β)

 

(b² - a² - c²) / -2ac = Cos(β)

 

Cos^-1[(b² - a² - c²) / -2ac] = β

 

 

To find the last angle, use the formula

 

α = 180 - (β + γ)