A triangle has sides A, B, and C. Sides A and B are of lengths #6# and #1#, respectively, and the angle between A and B is #(7pi)/8 #. What is the length of side C?

Question

A triangle has sides A, B, and C.  Sides A and B are of lengths #6# and #1#, respectively, and the angle between A and B is #(7pi)/8 #. What is the length of side C?

Emery Adkins · Accepted Answer

#C ~~ 6.93#

You can use the law of cosines here.

Let the angle opposite to side #A# be #alpha#, the angle opposite to side #B# be #beta# and the angle opposite to side #C# be #gamma#.

Then the law of cosines states:

#C^2 = A^2 + B^2 - 2ABcos(gamma)#

# = 6^2 + 1^2 - 2*6*1cos((7pi)/8)#

# = 37 - 12cos((7pi)/8)#

# ~~ 37 - 12 * (-0.92)#

# ~~ 48.09 " units"^2#

Thus,

#C ~~ 6.93 " units"#

Brooklyn Anderson · Answer

undefined To find the length of side C in the triangle, we can use the Law of Cosines, which states:

C² = A² + B² - 2AB * cos(C)

Given that sides A and B are of lengths 6 and 1 respectively, and the angle between them is (7π)/8, we can substitute these values into the formula:

C² = 6² + 1² - 2(6)(1) * cos(7π/8)

Solving for C:

C² = 36 + 1 - 12 * cos(7π/8)

Now, we need to find the cosine of (7π)/8. Once we have that value, we can continue solving for C.

Using a calculator or reference table, cos(7π/8) ≈ -0.9239

Substituting this value into the equation:

C² = 36 + 1 - 12 * (-0.9239)

C² = 37 + 11.0868

C² ≈ 48.0868

Taking the square root of both sides to solve for C:

C ≈ √48.0868

C ≈ 6.93

So, the length of side C is approximately 6.93.