find the area of the triangle with the given sides
B= 70 degrees 30' ; a=105 ; c=64
To find the area of a triangle with the given sides, we can use Heron's formula. Heron's formula states that the area (A) of a triangle with sides of lengths a, b, and c, where s is the semiperimeter (s = (a + b + c)/2), can be calculated using the formula:
A = √(s(s-a)(s-b)(s-c))
In this case, b is missing, but we can find it using the Law of Cosines. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and opposite angles A, B, and C, the following equation holds:
a^2 = b^2 + c^2 - 2bc * cos(A)
Plugging in the given values, we have:
A = 105
C = 64
B = 70 degrees 30'
To use the Law of Cosines, we need to convert B from degrees and minutes to just degrees. There are 60 minutes in a degree, so we have:
B = 70 + 30/60 = 70.5 degrees
Now we can rearrange the Law of Cosines formula to solve for b:
b^2 = a^2 + c^2 - 2ac * cos(B)
Plugging in the values, we have:
b^2 = 105^2 + 64^2 - 2(105)(64) * cos(70.5)
Now, we can use a calculator to evaluate this expression:
b^2 ≈ 11098.98
Taking the square root of both sides, we find:
b ≈ √(11098.98) ≈ 105.41
Now, we have all the side lengths (a = 105, b ≈ 105.41, and c = 64) to calculate the area using Heron's formula.
First, let's calculate the semiperimeter, s:
s = (a + b + c)/2 = (105 + 105.41 + 64)/2 = 167.71
Now, we can calculate the area using Heron's formula:
A = √(s(s-a)(s-b)(s-c)) = √(167.71(167.71-105)(167.71-105.41)(167.71-64))
Again, we can use a calculator to compute the value:
A ≈ √(167.71 * 62.71 * 62.3 * 103.71) ≈ √(663108898) ≈ 25718.05
Therefore, the area of the triangle with the given sides is approximately 25718.05 square units.