Asked by Tom
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.)
Answers
Answered by
Reiny
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
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
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