I need to calculate the area of the right angle triangle if we know the hypotenuse is 12...

... and one of the angles is 60 degrees.

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Given the hypotenuse in a right angle triangle is 12

Also, given that one of the angles is 60 degrees.

We need to find the area.

We will need to determine the length of the sides.

Let the sides be a , b, and c such that c is the hypotenuse.

==> A = (1/2) * a * b

Now we will use cos to find the sides.

We know that one of the sides = c*cos60

==> a = 12*cos60 = 12*1/2 = 6

Now we will determine the other side b.

==> b= 12*sin60 = 12*sqrt3/2 = 6sqrt3

Now we will substitute into the area.

==> A = (1/2)* a * b

= (1/2)*6 * 6sqrt3= 18sqrt3

**Then the area of the triangle is A = 18sqrt3 = 31.18 square units.**

The area could be calculated using the formula:

A = leg1*leg2*sin90/2

sin 60 = leg 1/hypotenuse => sqrt3/2 = leg1/12

leg 1 = 6sqrt3

cos 60 = leg2/12

1/2 = leg2/12

leg 2 = 6

The area is:

A = 6*6sqrt3/2

**A = 18*sqrt3 square units**

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