an isosceles triangle has sides of length 8cm, 8cm and 5cm. what is the angle between the equal sides

Question

This is for my math class cant seem to get it. dont know what to use to calculate the answer. please help someone.

Answer 1

draw the perpendicular bisector to the base whose measure is 5 cm

you have a right triangle whose hypotenuse is 8 and whose shorter leg is 2.5

using the Pythagorean Theorem

call the perpendicular bisector x

8^2=x^2+(2.5)^2

64=x^2+6.25

64-6.25=x^2

57.75=x^2

take the square root of both sides

x=7.59934 cm is the bisector

sin=opposite/hypotenuse

sinθ=7.59934/8

sinθ=0.9499175

use the arcsin

sin of what angle equals 0.9499175 ?

θ=71.78999º

double this angle and then subtract the sum from 180º

71.78999*2=143.57998º

180-143.57998=36.42º is the angle between the equal sides

this is longer than Daniel L's solution but if you can follow this you should be able to follow Daniel L's solution

2.5/8=0.3125

use the arccos to find the angle whose cosine is 0.3125

you should get approximately 71.79º and work from there

Answer 2

Hi Ash,

 

If you draw the triangle and drop a perpendicular down from the top it bisects the bottom side. 

 

Now you can use trigonometry to figure the angles.  The bottom angle adjacent side is 2.5 cm. long and the hypotenuse is 8 cm. long.  So you can find what angle has a cosine of 2.5/8.  The other bottom angle is the same.  Knowing that all angles have to add up to 180, you can calculate the top angle.

 

Have fun.

 

Dan