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°?

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.

[tex]\frac{x}{2.5}=\sin 12.5^\circ[/tex]

Multiply both sides by 2.5.
[tex]x=2.5(\sin 12.5^\circ)[/tex]

A calculator gives x = .541

The base is then double that value!

eco92 eco92 · Answer 2 · 2017-08-23T21:18:32+02:00

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:

[tex] |AB|^{2}= |CB|^{2}+ |CA|^{2}-2*|CB|*|CA|*cos(C)[/tex]

then substituting the values we know:

[tex] |AB|^{2}= (2.5)^{2}+ (2.5)^{2}-2*(2.5)*(2.5)*cos25[/tex]

[tex] |AB|^{2}= 2*(2.5)^{2}-2*(2.5)^{2}(0.906)[/tex]

[tex] |AB|^{2}= 2*(2.5)^{2}(1-0.906)[/tex]

[tex] |AB|^{2}= (2.5)^{2}(0.188)[/tex]

taking the square root of both sides:

[tex]|AB|= 2.5* \sqrt{0.2}=2.5*0.43358=1.08[/tex] (meters)

Answer: 1.08 m