isoceles triangle with vertex of 45 degrees, base angles are 67.5 degrees. The 2 legs are 10' in length. What is the length of the base? (I am an older adult with minimal math skill. Just have a building project.)
In your sketch , draw a perpendicular from the vertex to the base.
You now have 2 identical right-angled triangles.
let the base for each be x units (your whole base would then be 2x)
by simple trig,
cos 67.5° = x/10
x = 10cos67.5
= 3.8268.. , Use your calculator for this
so twice that is appr 7.654 '
which would be 7 ' and about 8 inches
or by the cosine law:
base^2 = 10^2 + 10^2 - 2(10)(10)cos 45
= 58.5786..
base = √58.5786
= appr 7.654 , same as above
To solve this problem, we can use the properties of an isosceles triangle. An isosceles triangle is a triangle with two congruent sides. In this case, you mentioned that the two legs of the triangle are both 10 feet in length.
First, let's label the triangle with side lengths and angles:
/\
/ \
10/ \10
/ \
/_______\
45°
We are given that the vertex angle (the angle at the top of the triangle) is 45 degrees, and the base angles (the angles at the bottom, where the legs meet the base) are both 67.5 degrees.
Since the sum of the angles in a triangle is always 180 degrees, we can calculate the measure of the base angle:
180 degrees = 45 degrees + 67.5 degrees + 67.5 degrees
180 degrees = 45 degrees + 135 degrees
180 degrees = 180 degrees
Therefore, the base angle is 180 degrees - 45 degrees - 67.5 degrees - 67.5 degrees = 180 degrees - 180 degrees = 0 degrees.
Now, an isosceles triangle with a base angle of 0 degrees is actually a degenerate triangle, which means the base has no length.
In conclusion, the length of the base of the isosceles triangle is 0 feet.