This diagram is a straightedge and compass construction. A is the center of one circle,
1 answer:
ABC is a equilateral triangle .
Proof :-
Let's assume both circles as C1 and C2 [ as shown in the figure ]
- AB is the radius of circle C1
- AB is the radius of Circle C2
AC is the radius of circle C1.
BC is the radius of circle C2 .
AB and AC both are radius of circle C1 so both are equal ie AB = AC .
AB and BC both are radius of circle C 2 so both are equal ie AB = BC .
Hence we conclude that .
AB = BC = AC.
So the triangle is equilateral triangle.
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t 1/2 =
t 1 = 1/2
t 2 = - 2 ( this solution is not acceptable )
sin x = 1/2
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x 1 = π / 6 + 2 kπ, x 2 = 5 π / 6 + 2 k π, k ∈ Z
3 sin x = 2 - 2 sin² x
2 sin² x + 3 sin x - 2 = 0
Substitution: t = sin x
2 t² + 3 t - 2 = 0
t 1/2 =
t 1 = 1/2
t 2 = - 2 ( this solution is not acceptable )
sin x = 1/2
Answer:
x 1 = π / 6 + 2 kπ, x 2 = 5 π / 6 + 2 k π, k ∈ Z