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sin[squared]60 is the same as (sin 60)^2 or (sin 60)*(sin 60)

If you dont know the sines and cosines of a 30-60-90 triangle off the top of your head, you can enter into a calculator to find out.

sin 60 = 3^(1/2)/2 OR "The square root of three divided by two"
cos 30 = 3^(1/2)/2 OR " ' ' ' ' ' ' ' ' "

So we end up with:

[ 3^(1/2) / 2 ]^2 + [ 3^(1/2) / 2 ]^2
Since it's the same thing twice, we can say:
2 * [ 3^(1/2) / 2 ]^2
Squaring...
2 * [ 3^(1/2) / 2 ] * [ 3^(1/2) / 2 ]
2 * [ 3 / 4]
[ 6 / 4 ] = 1.5

chandan
urs...........

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How do you construct a triangle with perimeter 150 mm and a base angle 75 degrees and 30 degrees?

Perhaps you can ask the angel to shed some divine light on the question! Suppose the base is BC, with angle B = 75 degrees angle C = 30 degrees then that angle A = 180 - (75+30) = 75 degrees. Suppose the side opposite angle A is of length a mm, the side opposite angle B is b mm and the side opposite angle C is c mm. Then by the sine rule a/sin(A) = b/(sin(B) = c/sin(C) This gives b = a*sin(B)/sin(A) and c = a*sin(C)/sin(A) Therefore, perimeter = 150 mm = a+b+c = a/sin(A) + a*sin(B)/sin(A) + a*sin(C)/sin(A) so 150 = a*{1/sin(A) + sin(B)/sin(A) + sin(C)/sin(A)} or 150 = a{x} where every term for x is known. This equation can be solved for a. So draw the base of length a. At one end, draw an angle of 75 degrees, at the other one of 30 degrees and that is it!


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Related Questions

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cos(30) = sqrt(3)/2 so cosine squared is 3/4.


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How do you construct a triangle with perimeter 150 mm and a base angle 75 degrees and 30 degrees?

Perhaps you can ask the angel to shed some divine light on the question! Suppose the base is BC, with angle B = 75 degrees angle C = 30 degrees then that angle A = 180 - (75+30) = 75 degrees. Suppose the side opposite angle A is of length a mm, the side opposite angle B is b mm and the side opposite angle C is c mm. Then by the sine rule a/sin(A) = b/(sin(B) = c/sin(C) This gives b = a*sin(B)/sin(A) and c = a*sin(C)/sin(A) Therefore, perimeter = 150 mm = a+b+c = a/sin(A) + a*sin(B)/sin(A) + a*sin(C)/sin(A) so 150 = a*{1/sin(A) + sin(B)/sin(A) + sin(C)/sin(A)} or 150 = a{x} where every term for x is known. This equation can be solved for a. So draw the base of length a. At one end, draw an angle of 75 degrees, at the other one of 30 degrees and that is it!


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