Michael W. answered • 12d

SW Engineer w/ MS CompSci & BS Math

Law of Sines: a/(sinA) = b/(sinB) = c/(sinC)

Law of Cosines: c^2 = a^2+b^2-2*a*b*cos(C)

A: Given two sides and an angle what combinations can this be a legit triangle

lets say the vertices are A, B, and C.

The sides are then AB, BC, CA, (for Law of Sines a = BC, b=CA, c = AB)

Suppose AB = 4cm, BC =5cm, CA = xcm (or a=5, b=x, c=4)

**First scenario: angle A = 50 degrees, angles B+C = 130 degrees**

5/sin(50) = 4/sin(C) solve for C

sin(C) * 5 /sin(50) = 4

sin(C) = 4*sin(50)/5

C = inverseSin(4*sin(50)/5) approximately 37 degrees

This means that B is approximately 93 degrees.

Now we solve for x (ie the length of CA)

5/sin(50) = x/sin(130-C)

sin(130-C)*5/sin(50) = x approximately 6.5cm

**The second scenario would be B = 50 degrees and A+C =130**

**and the third would be C = 50 degrees and A+B = 130**

**Second Scenario:**

Law of Sines: x/sin(50) = 4/sin(C) = 5/sin(A) = 5/sin(130-C)

Law of Cosines: b^2 = a^2+c^2-2*a*c*cos(B)

x^2 = 5^2+4^2-2*5*4*cos(50) solve for x

x=sqrt(25+16-40*cos(50)) approximately 3.9cm

x/sin(50) = 5/sin(A) solve for A

sin(A) * x /sin(50) = 5

sin(A) = 5*sin(50)/x

A = inverseSin(5*sin(50)/x) approximately 78 degrees

130-A = C approximately 52 degrees

**Third Scenario:**

4/sin(50) = 5/sin(A) solve for A

sin(A) * 4 /sin(50) = 5

sin(A) = 5*sin(50)/4

A = inverseSin(5*sin(50)/4) approximately 73 degrees

This means that B is approximately 57 degrees.

Now we solve for x (ie the length of CA)

4/sin(50) = x/sin(130-A)

sin(130-A)*4/sin(50) = x approximately 4.4cm

**Aside:**

I'm rounding, if you need to be more precise and are permitted the use of calculation programs, Excel works nicely for this. Just don't for get to convert the degrees to radians for the sin, cos functions and back to degrees on asin functions. SIN(RADIANS(50)) DEGREES(ASIN(xyx))

**For B:**

**A = 30 degrees; B = 60 degrees, C automatically = 90 degrees (triangle sum of angles =180)**

a=BC, b=CA, c=AB

Suppose a = 3cm

3/sin(30) = b/sin(60) = c/sin(90) solve for b and c

Suppose b = 3cm

a/sin(30) = 3/sin(60) = c/sin(90) solve for a and c

Suppose c = 3cm

a/sin(30) = b/sin(60) = 3/sin(90) solve for a and b