prove that sin 30°=1/2 or sin 60 =√3/2
Answers
Answer:
Step-by-step explanation:
Consider a 45-45-90 triangle. Because 2 angles are equal, it is
isosceles: both legs have the same length. If you know one leg (a),
you know both legs and you can use the Pythagorean Theorem to find the
hypotenuse c.
/|
/ |
c / |
/ | a
/ |
/ |
/____________|
a
c = sqrt(a^2 + a^2)
= sqrt(2*a^2)
= sqrt(2)*a
So these are the proportions of the sides of a 45-45-90 triangle:
B
/|
/ |
sqrt(2) / |
/ | 1
/ |
/ |
/____________|
A 1 C
In other words, the ratio of the hypotenuse to either leg is
sqrt(2):1. In terms of trigonometry,
sin(45) = sin(A)
= BC / AB
= 1/sqrt(2)
= sqrt(2)/2
Now for the 30-60-90 triangle. It isn't isosceles like the 45-45-90
triangle, but it is half of an equilateral triangle:
/|\
/ | \
c / | \ c
/ | \
/ a| \
/ | \
/______|______\
b b
The angles of the equilateral triangle are 60 degrees. The top angle
is bisected, giving the 30 degree angle. And since the sides of the
equilateral triangle are equal, 2b = c. That's the key - the side
opposite the 30 degree angle (b) is half the hypotenuse (c).
If we know b, then we know c = 2b, and we can fill in the Pythagorean
Theorem:
(2b)^2 = a^2 + b^2
a = sqrt((2b)^2 - b^2)
= sqrt(3b^2)
= sqrt(3)*b
Now we have the proportions of the sides of a 30-60-90 triangle:
B
/|
/ |
/ |
2 / |
/ | sqrt(3)
/ |
/______|
A 1 C
As before, we can evaluate the sine of angle A, which is 60 degrees:
sin(60) = BC / AB
= sqrt(3)/2
Likewise, for angle B:
sin(30) = AC / AB
= 1/2
Similar is the same
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