SOLUTION: sir, maam good day pls help me with my homework pls i need to answer this in 3 hour because i am confused so that i cannot do it on my house pls help me 1.Two altitudes of an

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Question 982458: sir, maam
good day pls help me with my homework pls i need to answer this in 3 hour because i am confused so that i cannot do it on my house pls help me
1.Two altitudes of an isosceles triangle areequal to 20 cm and 30 cm. Determine the baseangles of the triangle\
2.Two altitudes of an isosceles triangle areequal to 20 cm and 30 cm. Determine the baseangles of the triangle
Found 2 solutions by KMST, ikleyn:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a triangle is %281%2F2%29%28base%29%28altitude%29 .
The sides of an isosceles triangle come in two lengths:
the length of the two congruent legs, and
the length of the other side, which we usually consider the base.
However, you can take any side of a triangle as the base,
and the perpendicular to that base from the other vertex is the altitude,
so, in that case,
20cm is the altitude to a base measuring Bcm , and
30cm is the altitude to a base measuring bcm .
Calculating the area both ways, we get the same result, so
%281%2F2%29%2AB%2A20=%281%2F2%29%2Ab%2A30--->2B=3b--->B=%283%2F2%29b=1.5b .
Could the legs of the isosceles triangle measure Bcm ?
If so, the triangle looks like this
and cos%28red%28theta%29%29=%280.5b%29%2F%281.5b%29=0.5%2F1.5=1%2F3 ---> red%28theta%29=70.53%5Eo (rounded).
Could the legs of the isosceles triangle measure bcm ?
We would have an angle alpha such that
cos%28alpha%29=%280.5B%29%2Fb=0.5%2A%281.5b%29%2Fb=0.5%2A1.5=0.75--->alpha=41.41%5Eo (rounded).

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!

1.  First,  let us assume that the altitude of  20 cm  long goes  to the base  b  and the altitude of  30 cm  long goes to the lateral side  a.

Then you have these expressions for the area of the triangle:
S = 1%2F2.b%2A20 = 1%2F2.a%2A30.

It implies that  b%2A20 = a%2A30.

Hence,  b = a%2A%2830%2F20%29 = a%2A%283%2F2%29.

Now consider the right-angled triangle formed by the base,  by the altitude of  20 cm  long and by the lateral side  a.

This right-angled triangle has the hypotenuse of the length  a  and the leg of the length  b%2F2  (yes,  it is half of the base of the triangle,  b%2F2 = a%2A%283%2F4%29).

I believe  (I hope)  you just made a sketch of your triangle.  It will help you to see what I am saying.

Therefore,  cos%28alpha%29 = %28b%2F2%29%2Fa = %28a%2A%283%2F4%29%29%2Fa = 3%2F4,  where  alpha  is the angle at the base of our isosceles triangle.

Thus,  in this case cos%28alpha%29 = 3%2F4 and alpha = arccos%283%2F4%29.


2.  Next, let us consider the other configuration when the altitude of  30 cm  long goes  to the base  b  and the altitude of  20 cm  long goes to the lateral side  a.

Then you have these expressions for the area of the triangle:
S = 1%2F2.b%2A30 = 1%2F2.a%2A20.

It implies that  b%2A30 = a%2A20.

Hence,  b = a%2A%2820%2F30%29 = a%2A%282%2F3%29.

Now consider the right-angled triangle formed by the base,  by the altitude of  30 cm  long and by the lateral side  a.

This right-angled triangle has the hypotenuse of the length  a  and the leg of the length  b%2F2  (which is half of the base of the triangle, b%2F2 = a%2A%282%2F6%29) = a%2A%281%2F3%29.

Again, I hope you just made a sketch of your triangle.

Therefore,  cos%28alpha%29 = %28b%2F2%29%2Fa = %28a%2A%281%2F3%29%29%2Fa = 1%2F3.

Thus,  in this case cos%28alpha%29 = 1%2F3 and alpha = arccos%281%2F3%29.

Answer.  alpha = arccos%281%2F3%29  or  alpha = arccos%283%2F4%29.