can you help me with my maths homework?

Question

i'd do it myself, but i CANT.
in a triangle, the angle A = 71 degrees 48 minutes
|bc| = 18 cm
and |ac| = 12 cm
find i) (angle of) abc
ii) |ab| correct to the nearest cm

how do i do this??

Answer 1

side I BCI = a (side opposite to Angle A) = 18

side IACI = b = side opposite to angle B = 12

apply sin rule

a/Sin A = b/sin B

18/sin 71 = 12/sin B

sin B = (12/18) sin 71
= (3/4) sin 71
= 0.75(0.9455)
= 0.709

angle B = sin^-1(0.709)
B = 45.1 degrees
angle of ABC = 45.1 degrees

angle C = 180 -(A + B )
= 180 -(71 + 45)
= 180 - 116
= 64 degrees

a/sin A = c/Sin C

c = a sin C/sin A
= 18( sin 64)/sin 71
= 18((0.898/0.9455)
= 17 cm
IABI = 17 cm

note: usual notation of lines - capital letters, ex: line AB, angles -capital letters, ex angle A, triangle sides- small letters ex: side a

Answer 2

Well, the problem uses strange notation. In one place they use a capital letter (A) to stand for an angle; in another place they use 3 points (abc) to stand for an angle. Well, let's just use "B" to stand for angle abc, for consistency.

The Law of Sines says:

sin(A) / |bc| = sin(B) / |ac|

From that, you can figure out sin(B), and from that you can figure out Part 1.

Once you know the angles A and B, you can figure out the final angle C. You can do this because you know that A+B+C = 180° for any triangle.

Once you know angle C, you can use the Law of Cosines to figure out |ab|. According to the Law of Cosines:

|ab|² = |ac|² + |bc|² – 2(|ac|)(|bc|)cos(C)