Question 264868
The triangle is ABC


A is at the top, B is bottom left, C is bottom right.


BC is 12 cm.


Drop an altitude from point A intersecting with BC at point D.


The altitude is AD and it's length is 4 cm.


Let AB be the side that is 8 cm.


Let BC be the side that is 12 cm.


Sine of angle ABD = opposite / hypotenuse = 4/8 = 1/2.


angle ABD = arc-sine of 1/2 = 30 degrees.


Drop a perpendicular from point C intersecting with AB at point E.


This forms right triangle CEB.


This means that angle BCE is 60 degrees because angle EBC is the same as angle ABD which equals 30 degrees and angle BEC is 90 degrees and the sum of the angles of a triangle is 180 degrees.


sine of EBC = opposite / hypotenuse = EC / 12.


multiply both sides of this equation by 12 to get:


EC = 12 * sine of EBC = 12 * sine (30) = 6


this is the same as:


cosine of ECB = adjacent / hypotenuse = EC / 12.


multiply both sides of this equation by 12 to get:


EC = 12 * cosine of ECB = 12 * cosine (60) = 6


Answer is that the altitude to the 8 cm side is equal to 6.


That would be selection D.


A picture of your triangle is shown below.


<img src = "http://theo.x10hosting.com/problems/264868.jpg" height = "300" />


In this diagram:

AB = 8
BC = 12
AD = 4
EC = 6
angle ABD and ABC and EBC are the same angle that equals 30 degrees.
Angle BEC = 90 degrees
Angle ECB = 60 degrees