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How Dividing equilateral triangle? - Answers

This is a fun question. If the triangle rests on one of its sides as a base, then the altitude of the triangle, a line drawn from the apex of the triangle to the base, divides the triangle into two right angle 30o,60o, 90o, triangles. For convenience let each of the equal sides of the triangle 2 units. Then Pythagorus tells us that the base of each of these right triangles is 1 unit and the altitude is √3. This leads directly to sin(30o) cos(60o) 1/2 and cos(30o) sin(60o) √3/2 0.866 rounded. You can also use one of these right angle triangles to find the sine and cosine of 15o but the algebra gets a little messy.



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How Dividing equilateral triangle? - Answers

https://math.answers.com/geometry/How_Dividing_equilateral_triangle

This is a fun question. If the triangle rests on one of its sides as a base, then the altitude of the triangle, a line drawn from the apex of the triangle to the base, divides the triangle into two right angle 30o,60o, 90o, triangles. For convenience let each of the equal sides of the triangle 2 units. Then Pythagorus tells us that the base of each of these right triangles is 1 unit and the altitude is √3. This leads directly to sin(30o) cos(60o) 1/2 and cos(30o) sin(60o) √3/2 0.866 rounded. You can also use one of these right angle triangles to find the sine and cosine of 15o but the algebra gets a little messy.



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https://math.answers.com/geometry/How_Dividing_equilateral_triangle

How Dividing equilateral triangle? - Answers

This is a fun question. If the triangle rests on one of its sides as a base, then the altitude of the triangle, a line drawn from the apex of the triangle to the base, divides the triangle into two right angle 30o,60o, 90o, triangles. For convenience let each of the equal sides of the triangle 2 units. Then Pythagorus tells us that the base of each of these right triangles is 1 unit and the altitude is √3. This leads directly to sin(30o) cos(60o) 1/2 and cos(30o) sin(60o) √3/2 0.866 rounded. You can also use one of these right angle triangles to find the sine and cosine of 15o but the algebra gets a little messy.

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      This is a fun question. If the triangle rests on one of its sides as a base, then the altitude of the triangle, a line drawn from the apex of the triangle to the base, divides the triangle into two right angle 30o,60o, 90o, triangles. For convenience let each of the equal sides of the triangle 2 units. Then Pythagorus tells us that the base of each of these right triangles is 1 unit and the altitude is √3. This leads directly to sin(30o) cos(60o) 1/2 and cos(30o) sin(60o) √3/2 0.866 rounded. You can also use one of these right angle triangles to find the sine and cosine of 15o but the algebra gets a little messy.

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