side AC= 16
c=30 degrees
b=90 degrees
a=60 degrees
Solve for BA rounded to the nearest tenth =
Solve for BC rounded to the nearest tenth =
To solve for BA, we can use the sine rule. The sine rule states that for a triangle with sides a, b, and c and their opposite angles A, B, and C respectively:
a/sin(A) = b/sin(B) = c/sin(C)
To solve for BA, we have:
BA/sin(B) = AC/sin(A)
BA/sin(90) = 16/sin(60)
BA/1 = 16/(√3/2)
BA = 16√3/2
BA ≈ 13.9 (rounded to the nearest tenth)
To solve for BC, we can use the cosine rule. The cosine rule states that for a triangle with sides a, b, and c and the angle C between sides a and b:
c² = a² + b² - 2abcos(C)
Substituting the given values:
BC² = BA² + AC² - 2(BA)(AC)cos(b)
BC² = (13.9)² + 16² - 2(13.9)(16)cos(30)
BC² ≈ 192.4 + 256 - 446.4(0.866)
BC² ≈ 192.4 + 256 - 386.6
BC² ≈ 61.8
Taking the square root:
BC ≈ √61.8
BC ≈ 7.9 (rounded to the nearest tenth)