Thread: Isoseles triangle

What is the measure of each base angle of an isosceles triangle if its vertex angle measures 42 degrees and its 2 congruent sides measure 12 units?

How would you solve this problem i am completely lost. Please explain and give the answer,

Thanks A Lot!

Originally Posted by OnMyWayToBeAMathProffesor

What is the measure of each base angle of an isosceles triangle if its vertex angle measures 42 degrees and its 2 congruent sides measure 12 units?

How would you solve this problem i am completely lost. Please explain and give the answer,

Thanks A Lot!

We have two sides and the angle inbetween.

We can drop a perpendicular bisector in the triangle, splitting the vertex angle up into 21 degrees and 21 degrees. (And thus we have two right triangles).

Using Law of Sines:

(Sin90)/12 = (Sin21)/b

Solve for b:

b = 11.2303

Thus, the total base is 22.4606

Nevermind, I just realized you wanted to find just the base angles; this is easy from just subtracting each angle 180 - 21 - 90 = 69 degrees, and obviously each of the base angles will be congruent since it's an isosceles triangle.

thanxs a lot, i get it now