What is the approximate length of the base of an isosceles triangle if the congruent sides are 2.5 m and the vertex angle is 25°?

Let CH be the altitude.

CH is also the angle bisector of angle C, so:

m(ACH)=m(HCB)=25°/2=12.5° 

CH is also a median, so |AH|=|HB|

Method 1:

by right angle trigonometry, in triangle HBC

|HB|=|CB|*sin12.5° (as sine = opposite side / hypotenuse) 

|HB|= 2.5 * 0.216 = 0.54 (meters)

thus, |AB|=2|HB|=2*0.54 m = 1.08 m

Method 2:

according to the Cosine law:



then substituting the values we know:









taking the square root of both sides:

  (meters)

1.08 m

I'm assuming you know the basic trigonometric functions in a right triangle:

sine = opposite / hypotenuse,  cosine = adjacent / hypotenuse
tangent = opposite / adjacent

See attached diagram.  Draw a segment perpendicular to the base forming a right triangle and dividing the base into 2 equal parts.

The apex angle is split in half, so 12.5 degrees in each half.

x is the opposite side, 2.5 m is the hypotenuse, so use the sine ratio.



Multiply both sides by 2.5.


A calculator gives  x = .541

The base is then double that value!

The third box is a function