problem:
A field in the shape of a triangle, ABC. It is found that :
side AB is 580 m,
angle A is 51.3
angle B is 58.2
what is the perimeter of the field????
2007-03-30 11:56:18 · 3 answers · asked by mathelp 2 in Science & Mathematics ➔ Mathematics
well if you use "A" to represent the angles "a" to represent the sides
then AB equall "c" because its the side opposite
A =51.3 a
B =58.2 b
C =70.5 (because total =180 c = 580
use
a/sinA = b/sinB = c/sinC
580/sin70.5 = b/sin58.2 (calc is in degree mode)
b= 522.932
580/sin70.5= a/sin51.3
a=480.1927
318.50 m ( make sure when you add a+b + 580 you use second enter in the calc to be more precise)
2007-03-30 12:13:59 · answer #1 · answered by acta non verba 3 · 0⤊ 0⤋
use the law of sines:
a/sin(A) = b/sin(B) = c/sin(C)
where a, b, and c are the lengths of the sides opposite angles at A, B, and C resply.
you know side AB (which is side c)
angle C can be calculated as 180 - A -B
from this you can get a, b, and c and add them.
2007-03-30 12:01:17 · answer #2 · answered by astatine 5 · 0⤊ 0⤋
angle C is 180-(58.2+51.3) = 70.5
Now you can use the law of sines:
580/sin 70.5 = AC/sin 58.2 = BC/sin 51.3
2007-03-30 12:08:17 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋