select all the conditions it is possible to construct a triangle. and why?

Question

select all the conditions it is possible to construct a triangle. and why?

a. triangle with side length 4cm,5cm,and 50 degrees angle?

b. triangle with angle measure 30 and 60 degrees and 3cm side length?

Answer 1

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