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Points A, B and C on a circle of radius r are situated so that AB = AC, AB > r, and the length of minor arc BC is r.
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Points A, B and C on a circle of radius r are situated so that AB = AC, AB > r, and the length of minor arc BC is r. If angles are measured in radians, then AB/BC = ?
A) 1/2 csc(1/4)
B) 2 cos(1/2)
C) 4 sin(1/2)
D) csc(1/2)
E) 2 sec(1/2)
So, I got the lengths of AB and BC in terms of the radius using law of cosines.
-> AB = sqrt(2r^2(1+cos(1/2)) and BC = sqrt(2r^2(1-cos(1))
I put them in a fraction and simplified using the double angle formula to get
-> AB/BC = 1/sqrt(2(1+cos(1/2))
From then, I don't know what to do since all the answer choices don't have a square root.
I'd appreciate some help, thanks!
trigonometry contest-math circle
asked Aug 1 at 0:13
ShadyAF
288
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up vote
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Points A, B and C on a circle of radius r are situated so that AB = AC, AB > r, and the length of minor arc BC is r. If angles are measured in radians, then AB/BC = ?
A) 1/2 csc(1/4)
B) 2 cos(1/2)
C) 4 sin(1/2)
D) csc(1/2)
E) 2 sec(1/2)
So, I got the lengths of AB and BC in terms of the radius using law of cosines.
-> AB = sqrt(2r^2(1+cos(1/2)) and BC = sqrt(2r^2(1-cos(1))
I put them in a fraction and simplified using the double angle formula to get
-> AB/BC = 1/sqrt(2(1+cos(1/2))
From then, I don't know what to do since all the answer choices don't have a square root.
I'd appreciate some help, thanks!
trigonometry contest-math circle
asked Aug 1 at 0:13
ShadyAF
288
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Points A, B and C on a circle of radius r are situated so that AB = AC, AB > r, and the length of minor arc BC is r. If angles are measured in radians, then AB/BC = ?
A) 1/2 csc(1/4)
B) 2 cos(1/2)
C) 4 sin(1/2)
D) csc(1/2)
E) 2 sec(1/2)
So, I got the lengths of AB and BC in terms of the radius using law of cosines.
-> AB = sqrt(2r^2(1+cos(1/2)) and BC = sqrt(2r^2(1-cos(1))
I put them in a fraction and simplified using the double angle formula to get
-> AB/BC = 1/sqrt(2(1+cos(1/2))
From then, I don't know what to do since all the answer choices don't have a square root.
I'd appreciate some help, thanks!
trigonometry contest-math circle
asked Aug 1 at 0:13
ShadyAF
288
Points A, B and C on a circle of radius r are situated so that AB = AC, AB > r, and the length of minor arc BC is r. If angles are measured in radians, then AB/BC = ?
A) 1/2 csc(1/4)
B) 2 cos(1/2)
C) 4 sin(1/2)
D) csc(1/2)
E) 2 sec(1/2)
So, I got the lengths of AB and BC in terms of the radius using law of cosines.
-> AB = sqrt(2r^2(1+cos(1/2)) and BC = sqrt(2r^2(1-cos(1))
I put them in a fraction and simplified using the double angle formula to get
-> AB/BC = 1/sqrt(2(1+cos(1/2))
From then, I don't know what to do since all the answer choices don't have a square root.
I'd appreciate some help, thanks!
trigonometry contest-math circle
asked Aug 1 at 0:13
ShadyAF
288
asked Aug 1 at 0:13
ShadyAF
288
asked Aug 1 at 0:13
ShadyAF
288
asked Aug 1 at 0:13
ShadyAF
288
288
add a comment |Â
add a comment |Â
1 Answer
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For your answer,
$$
frac 1 sqrt2(1 + cos (1/2)) = frac 1 sqrt 2 cdot 2 cos^2(1/4) = frac 12 sec (1/4),
$$
so perhaps you made some mistake during computation.
Now I provide my solution.
Solution. $blacktriangleleft$ Let the center of the circle be $O$. Since the length of the minor arc $BC$ is $r$, the angle $$angle BOC = r/r = 1,$$ which means $angle BAC = 1/2$. Now construct $AD perp BC$ where $D$ is on $BC$. Since $AB=AC$, the line segment $AD$ bisect $angle BAC$ and $BC$ simultaneously. Now in the right triangle $triangle ABD$, we have $$angle BAD = angle BAC /2 =1/4,$$ hence $$BD/AB = sin (angle BAD) = sin (1/4).$$ Therefore $$BC/AB = 2BD/AB = 2sin (1/4),$$ or $AB/BC = csc(1/4) /2$, which is exactly choice $(mathrm A)$. $blacktriangleright$
answered Aug 1 at 3:26
xbh
1,0257
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
For your answer,
$$
frac 1 sqrt2(1 + cos (1/2)) = frac 1 sqrt 2 cdot 2 cos^2(1/4) = frac 12 sec (1/4),
$$
so perhaps you made some mistake during computation.
Now I provide my solution.
Solution. $blacktriangleleft$ Let the center of the circle be $O$. Since the length of the minor arc $BC$ is $r$, the angle $$angle BOC = r/r = 1,$$ which means $angle BAC = 1/2$. Now construct $AD perp BC$ where $D$ is on $BC$. Since $AB=AC$, the line segment $AD$ bisect $angle BAC$ and $BC$ simultaneously. Now in the right triangle $triangle ABD$, we have $$angle BAD = angle BAC /2 =1/4,$$ hence $$BD/AB = sin (angle BAD) = sin (1/4).$$ Therefore $$BC/AB = 2BD/AB = 2sin (1/4),$$ or $AB/BC = csc(1/4) /2$, which is exactly choice $(mathrm A)$. $blacktriangleright$
answered Aug 1 at 3:26
xbh
1,0257
add a comment |Â
up vote
0
down vote
accepted
For your answer,
$$
frac 1 sqrt2(1 + cos (1/2)) = frac 1 sqrt 2 cdot 2 cos^2(1/4) = frac 12 sec (1/4),
$$
so perhaps you made some mistake during computation.
Now I provide my solution.
Solution. $blacktriangleleft$ Let the center of the circle be $O$. Since the length of the minor arc $BC$ is $r$, the angle $$angle BOC = r/r = 1,$$ which means $angle BAC = 1/2$. Now construct $AD perp BC$ where $D$ is on $BC$. Since $AB=AC$, the line segment $AD$ bisect $angle BAC$ and $BC$ simultaneously. Now in the right triangle $triangle ABD$, we have $$angle BAD = angle BAC /2 =1/4,$$ hence $$BD/AB = sin (angle BAD) = sin (1/4).$$ Therefore $$BC/AB = 2BD/AB = 2sin (1/4),$$ or $AB/BC = csc(1/4) /2$, which is exactly choice $(mathrm A)$. $blacktriangleright$
answered Aug 1 at 3:26
xbh
1,0257
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
For your answer,
$$
frac 1 sqrt2(1 + cos (1/2)) = frac 1 sqrt 2 cdot 2 cos^2(1/4) = frac 12 sec (1/4),
$$
so perhaps you made some mistake during computation.
Now I provide my solution.
Solution. $blacktriangleleft$ Let the center of the circle be $O$. Since the length of the minor arc $BC$ is $r$, the angle $$angle BOC = r/r = 1,$$ which means $angle BAC = 1/2$. Now construct $AD perp BC$ where $D$ is on $BC$. Since $AB=AC$, the line segment $AD$ bisect $angle BAC$ and $BC$ simultaneously. Now in the right triangle $triangle ABD$, we have $$angle BAD = angle BAC /2 =1/4,$$ hence $$BD/AB = sin (angle BAD) = sin (1/4).$$ Therefore $$BC/AB = 2BD/AB = 2sin (1/4),$$ or $AB/BC = csc(1/4) /2$, which is exactly choice $(mathrm A)$. $blacktriangleright$
answered Aug 1 at 3:26
xbh
1,0257
For your answer,
$$
frac 1 sqrt2(1 + cos (1/2)) = frac 1 sqrt 2 cdot 2 cos^2(1/4) = frac 12 sec (1/4),
$$
so perhaps you made some mistake during computation.
Now I provide my solution.
Solution. $blacktriangleleft$ Let the center of the circle be $O$. Since the length of the minor arc $BC$ is $r$, the angle $$angle BOC = r/r = 1,$$ which means $angle BAC = 1/2$. Now construct $AD perp BC$ where $D$ is on $BC$. Since $AB=AC$, the line segment $AD$ bisect $angle BAC$ and $BC$ simultaneously. Now in the right triangle $triangle ABD$, we have $$angle BAD = angle BAC /2 =1/4,$$ hence $$BD/AB = sin (angle BAD) = sin (1/4).$$ Therefore $$BC/AB = 2BD/AB = 2sin (1/4),$$ or $AB/BC = csc(1/4) /2$, which is exactly choice $(mathrm A)$. $blacktriangleright$
answered Aug 1 at 3:26
xbh
1,0257
edited Aug 1 at 4:02
answered Aug 1 at 3:26
xbh
1,0257
answered Aug 1 at 3:26
xbh
1,0257
answered Aug 1 at 3:26
xbh
1,0257
1,0257
add a comment |Â
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