please help, i need a complete solution and PLEASE WITH ILLUSTRATION of the triangle(DRAWING) dont make it in sentence 1. two altitudes of an isosceles triangle are equal to 20cm and 30cm. determine the possible measures of the the base angles of the triangle Solution Consider an isosceles triangle ABC with AB = BC. It may be AC < AB or AC > AB - first draw it so AB > AC. Then the altitude from M to B is 30. The other altitude is from C to D and that is 20. (It can not have AC = AB or all the sides would be equal and the altitudes would also be equal - you may wish to prove this for yourself.) Construct the perpendicular to AB through point C and let D be the intersection, so D is on AB and ADC and BDC are right angles. See that triangle ABM is similar to triangle CAD. (You may wish to prove this in detail.) Therefore AC/MB = hypotenuse/hypotenuse = longer leg / longer leg = 20 / 30. (If the altitudes are reversed so the distance from the unequal side to the opposite vertex is 20 instead of 30 then this is 30/20 instead.) Since AM = 1/2 AC this shows AM / MB = 10/30 (in the other case it is 15/20 instead.) Which trigonometric function can we use? The tangent of BAM is the opposite side (BM) over the adjacent side (AM) to the angle (BAM with A at the vertex). BM / AM = 30 / 10 = 3 What angle a has tan a = 3? This can be solved with a calculator, using the inverse tangent function or arctan function. Or it can be found using a trigonometric table of trig values or some other methods of computation that are also good to know about but too much for me to explain here. arctan(3) = CAB = 1.249 radians = 71.6 degrees arctan(4/3) = CAB = 0.927 radians = 53.1 degrees.

1. please help, i need a complete solution and PLEASE WITH ILLUSTRATION of the
triangle(DRAWING) dont make it in sentence
1. two altitudes of an isosceles triangle are equal to 20cm and 30cm. determine the possible
measures of the the base angles of the triangle
Solution
Consider an isosceles triangle ABC with AB = BC.
It may be AC < AB or AC > AB - first draw it so AB > AC. Then the altitude from M to B is 30.
The other altitude is from C to D and that is 20. (It can not have AC = AB or all the sides would
be equal and the altitudes would also be equal - you may wish to prove this for yourself.)
Construct the perpendicular to AB through point C and let D be the intersection, so D is on AB
and ADC and BDC are right angles.
See that triangle ABM is similar to triangle CAD. (You may wish to prove this in detail.)
Therefore AC/MB = hypotenuse/hypotenuse = longer leg / longer leg = 20 / 30. (If the altitudes
are reversed so the distance from the unequal side to the opposite vertex is 20 instead of 30 then
this is 30/20 instead.)
Since AM = 1/2 AC this shows AM / MB = 10/30 (in the other case it is 15/20 instead.)
Which trigonometric function can we use? The tangent of BAM is the opposite side (BM) over
the adjacent side (AM) to the angle (BAM with A at the vertex).
BM / AM = 30 / 10 = 3
What angle a has tan a = 3?
This can be solved with a calculator, using the inverse tangent function or arctan function.
Or it can be found using a trigonometric table of trig values or some other methods of
computation that are also good to know about but too much for me to explain here.

2. arctan(3) = CAB = 1.249 radians = 71.6 degrees
arctan(4/3) = CAB = 0.927 radians = 53.1 degrees