Question

The perimeter ΔABC is 30 feet. AB = 5x-7; BC = 3x + 1; AC= 4x. What are the angles in this triangle?

Answer 1

`A ~~ 55.7714^@, B~~82.8190^@, C~~41.4096^@`

Explanation:

Perimeter: `(5x-7) + (3x+1 )+(4x) = 30`

Combine like-terms and solve for `x`: `12x -6 = 30`
`12x = 36; x = 36/12 = 3` feet

Side lengths: `AB = 8` ft; `BC = 10` ft and `AC = 12` ft

Use the Law of Cosines to find one angle,
let `AB = c`, `BC=a`, and `AC = b`

`C^@ = arccos((a^2+b^2-c^2)/(2ab)) = arccos ((10^2+12^2-8^2)/(2*10*12)) = arccos (180/240) = arccos(3/4) ~~41.4096^@`

Use the Law of Sines to find the second angle B:
`b/(sinB) = c/(sinC)`

Using the cross product: `c sinB = b sinC`

Solve for angle B:

`B^@ = arcsin((b sinC)/c) = arcsin((12sin(41.4096))/8) ~~82.8190^@`

Find the third angle by Third Angle Sum Theorem: `A^@+B^@+C^@ = 180^@`

`A^@ = 180^@ - 41.4096^@ - 82.8190^@ = 55.7714^@`