Question

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°?

Answer 1

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.


`(x)/(2.5)=\sin 12.5^\circ`

Multiply both sides by 2.5.

`x=2.5(\sin 12.5^\circ)`

A calculator gives x = .541

The base is then double that value!

Answer 2

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:


`|AB|^(2)= |CB|^(2)+ |CA|^(2)-2*|CB|*|CA|*cos(C)`

then substituting the values we know:


`|AB|^(2)= (2.5)^(2)+ (2.5)^(2)-2*(2.5)*(2.5)*cos25`


`|AB|^(2)= 2*(2.5)^(2)-2*(2.5)^(2)(0.906)`




`|AB|^(2)= 2*(2.5)^(2)(1-0.906)`


`|AB|^(2)= (2.5)^(2)(0.188)`

taking the square root of both sides:


`|AB|= 2.5* √(0.2)=2.5*0.43358=1.08` (meters)


Answer: 1.08 m