SCALENE TRIANGLE

[moy2414]

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10A-40

WHY WONT IT WORK WHEN WE DO THE EQUATION IN THE FOLLOWING MANNER:

since we are working in Radians: 180º = ∏ and a straight angle = 180º then ∏-2.34 = .80159265359

then by using the law of sines: sin(.80159265359)/.0348 = sin(x)/.0476 ➔ x = 1.38466601298

then all angles of a triangle = 180º so ∏-(.80159265359+1.38466601298) = .95533398702 ➔ 9.55X10⁻¹ but the answer is 5.83X10⁻¹



BUT IF YOU WORK IT OUT IN THE FOLLOWING MANNER:

c² = a² + b² + 2ab(cosC') ➔ (.0348)² = (.0476)² + b² + 2(.0348)(b)(cos(2.34))

the variable b(this is the third side length) = .02666882

Now using the formula with the new data:

c² = a² + b² - 2ab(cosC) ➔ (.026668821)² = (.0476)² + (.0348)² - 2(.0348)(.0476)(cosC)

the answer to the problem is 5.8307352458X10⁻¹


WHY DOESN'T THE FIRST METHOD WORK?

[bradp]

1) stop typing in all caps

2) Google "SSA triangle the ambiguous case"
sin(.80159265359)/.0348 = sin(x)/.0476 ➔ x = 1.38466601298 ----> x can also be pi minus 1.384 = 1.76 because sin(180-x) = sin(x) when you're working from 0 degrees to 180 degrees. 1.76 leads to the answer you were looking for


Because a*sin(B)<b<a and B is acute, the triangle has two solutions. I guess the obtuse angle on the top suggests that the smaller angle is the answer.