Consider the triangle ABC. Suppose that a=73 b=71 c=3 Solve the triangle. Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.

Question

I NEED HELP WITH THIS ASAP PLEASE!!!

The triangle is ABC

c between AB

a between BC

b between AC

Answer 1

C=about 1.8 degrees = angle ACB or just C

B= 47.3 degrees = angle ABC or just B

A = 130.9 degrees = angle BAC or just A

Use the law of cosines. It's similar to the Pythagorean Theorem but with an extra term -2abCosC

c^2 = a^2 + b^2 -2abCosC

a=73, b=71, c= 3

CosC = 10361/10366=

0.9995 use the inverse trig function on a calculator

C = 1.77796= about 1.8 dgrees

use the law of cosines again to find B

b^2 = a^2 + c^2 -2acCosB

71^2 = 73^2 + 3^2 -2(73)(3)CosB

CosB = 297/438 - .,6708

= about 47.3060

then find A= 180-B-C = 180-1.8-47.3- 1.8 = 130.9

check that by using law of cosines again, or law of sines

a^2 = b^2 + c^2 -2bcCosA

73^2 = 71^2+ 3^2 - 2(71)(3)cosA

CosA = -279/426=-.6549

A = 130.9

sinA/a = sinC/c

sinA = asinC/c = 73sin1.77796/3 = 49.0230

A = 180-49.023 = 130.977 = 130.9

that's the basic method(s). Use law of cosines once, and then either use it one or two more times, or try the law sines and make sure the 3 angles sum to 180. Helps to sketch a rough picture of the triangle. A, B, C = about 130.9, 47.3, and 1.8 degrees

first observation is that any side of a triangle must be greater than the difference of the other 2 sides

3> 73-71, but not by much so angle opposite 3 will be near zero. the side opposite 73 will be the largest angle and the side opposite 71 will be the middle sized angle.

the 71 and 73 are so similar, in tedious calculations it's easy to mix them up and an error to creep in.