construct a right angled triangle whose perimeter is 10cm and one acute angle is equal to 60?

For a right triangle with one angle 60, the other angle is 30.
This is known to be half of a triangle with all corners 60 and all sides equal.

If the hypotenuse has length x, the short side has length x/2.
This makes the other side have length sqrt(x - x/4) = x*sqrt(3)/2.
The sum is then x + x/2 + x*sqrt(3)/2 = [1.5 + sqrt(3)/2]x.

Since we are told [1.5 + sqrt(3)/2]x = 10cm.
Multiplying the 1.5 by 2/2 and combining with the other terms gives
([3 + sqrt(3)]/2)x = 10.

Multiplying both side by 2 gives (3 + sqrt(3))x = 20.
Multiplying both sides by the conjugate of 3 + sqrt(3) gives (3 - 3)x = 20(3 - sqrt(3)).
Since 3-3 = 9-3 = 6, divide both sides by 6 and get x = 20(3 - sqrt(3))/6.
Since 20 is 2*10 and 6 is 2*3, that reduces to 10(3 - sqrt(3))/3.
It could also be split into two terms, giving 10 - sqrt(3))/3.


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Scott A Wilson


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