How do you construct a triangle with perimeter 150 mm and a base angle 75 degrees and 30 degrees?

Answer 1

Perhaps you can ask the angel to shed some divine light on the question!

Suppose the base is BC, with

angle B = 75 degrees

angle C = 30 degrees

then that angle A = 180 - (75+30) = 75 degrees.

Suppose the side opposite angle A is of length a mm, the side opposite angle B is b mm and the side opposite angle C is c mm.

Then by the sine rule

a/sin(A) = b/(sin(B) = c/sin(C)

This gives b = a*sin(B)/sin(A)

and c = a*sin(C)/sin(A)

Therefore, perimeter = 150 mm = a+b+c = a/sin(A) + a*sin(B)/sin(A) + a*sin(C)/sin(A)

so 150 = a*{1/sin(A) + sin(B)/sin(A) + sin(C)/sin(A)}

or 150 = a{x}

where every term for x is known.

This equation can be solved for a.

So draw the base of length a. At one end, draw an angle of 75 degrees, at the other one of 30 degrees and that is it!