finding altitude in triangles?

Question

All the problem gives me is the vertex=70 degrees, and the base=246, and it wants me to find the triangles altitude. It is an isosceles triangle. How am i supposed to figure this out?

Answer 1

If it is isosceles, then the two base angles are the same, and since they equal 180-70, they are both 55 degrees.

tan(55) = alt/(.5 * 246) = alt/123

alt = 123 * tan(55)
= 123 * 1.428 = 175.66

Answer 2

Because you have an isosceles triangle, the altitude is the distance from the point where the two equal sides meet to the base. If you drop a line from that point to the base, it will create a right angle and bisect the base. Now, you will have two identical right triangles inside the isosceles triangle, each with a base of 246/2= 123

OK. Now, we know that a triangle always has 180 degrees in it. So if we say that a base angle is x, and that both base angles have the same measure, then 2x+70=180

or 2x=110 or x=55

So, since we have a right triangle, we can calculate the altitude as tan55=a/123 where a is the altitude.

So, 1.43 = a/123

a=123*1.43 = 175.7

Answer 3

The altitude from the vertex angle of an isosceles triangle is also an angle bisector and a median (I tell my students you get "3 for the price of 1!"). So the altitude will bisect the 70° into two 35° angles. It will obviously be perpendicular to the base, which it bisects into two segments of length 123. So you use a tangent ratio to solve for the altitude, a. Tan 35° = 123/a --> a = 123/tan35° = 175.66