Trig and Triangle Math Club Question: Finding Side Length
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I recently had a math club competition, and I was unsure of how to approach one of the problems on the test:
In $triangle ABC$,
$ cos(2A-B) + sin(A + B) = 2$
$ overline{AB} = 4$.
What is the length of $overline{BC}$?
My immediate reaction was to simplify down all the trig functions, which resulted in: $$(cos^2 A - sin^2 A)cos B + 2sin A cos A sin B + sin A cos B + sin B cos A = 2$$
And I don't see any way to further simplify it. I think my original method is wrong, or I just can't see the next step forward, so I would really appreciate any tips on how to solve the problem.
trigonometry triangle recreational-mathematics problem-solving
asked Nov 15 at 6:23
Christopher Marley
86815
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up vote
3
down vote
favorite
I recently had a math club competition, and I was unsure of how to approach one of the problems on the test:
In $triangle ABC$,
$ cos(2A-B) + sin(A + B) = 2$
$ overline{AB} = 4$.
What is the length of $overline{BC}$?
My immediate reaction was to simplify down all the trig functions, which resulted in: $$(cos^2 A - sin^2 A)cos B + 2sin A cos A sin B + sin A cos B + sin B cos A = 2$$
And I don't see any way to further simplify it. I think my original method is wrong, or I just can't see the next step forward, so I would really appreciate any tips on how to solve the problem.
trigonometry triangle recreational-mathematics problem-solving
asked Nov 15 at 6:23
Christopher Marley
86815
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I recently had a math club competition, and I was unsure of how to approach one of the problems on the test:
In $triangle ABC$,
$ cos(2A-B) + sin(A + B) = 2$
$ overline{AB} = 4$.
What is the length of $overline{BC}$?
My immediate reaction was to simplify down all the trig functions, which resulted in: $$(cos^2 A - sin^2 A)cos B + 2sin A cos A sin B + sin A cos B + sin B cos A = 2$$
And I don't see any way to further simplify it. I think my original method is wrong, or I just can't see the next step forward, so I would really appreciate any tips on how to solve the problem.
trigonometry triangle recreational-mathematics problem-solving
asked Nov 15 at 6:23
Christopher Marley
86815
I recently had a math club competition, and I was unsure of how to approach one of the problems on the test:
In $triangle ABC$,
$ cos(2A-B) + sin(A + B) = 2$
$ overline{AB} = 4$.
What is the length of $overline{BC}$?
My immediate reaction was to simplify down all the trig functions, which resulted in: $$(cos^2 A - sin^2 A)cos B + 2sin A cos A sin B + sin A cos B + sin B cos A = 2$$
And I don't see any way to further simplify it. I think my original method is wrong, or I just can't see the next step forward, so I would really appreciate any tips on how to solve the problem.
trigonometry triangle recreational-mathematics problem-solving
trigonometry triangle recreational-mathematics problem-solving
asked Nov 15 at 6:23
Christopher Marley
86815
asked Nov 15 at 6:23
Christopher Marley
86815
asked Nov 15 at 6:23
Christopher Marley
86815
asked Nov 15 at 6:23
Christopher Marley
86815
asked Nov 15 at 6:23
Christopher Marley
86815
86815
add a comment |
add a comment |
2 Answers
2
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oldest
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up vote
3
down vote
accepted
Since
$$cos(2A-B)le 1 text{and} sin(A + B)le 1;\
cos(2A-B) = sin(A + B) = 1 Rightarrow \
begin{cases}2A-B=0 \
A+B=frac{pi}{2} end{cases}Rightarrow begin{cases}A=frac{pi}{6} \ B=frac{pi}{3}end{cases} Rightarrow C=frac{pi}{2}.$$
Hence:
$$BC=ABsin frac{pi}{6}=2.$$
answered Nov 15 at 6:42
farruhota
17.7k2736
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up vote
4
down vote
Hint:
For real $x,$ $$cos x,sin xle1$$
$$cos(2A-B) + sin(A + B) = 2impliescos(2A-B)=sin(A+B)=1$$
and $sin(A+B)=sin(pi-C)=sin C$
Can you take it from here?
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Since
$$cos(2A-B)le 1 text{and} sin(A + B)le 1;\
cos(2A-B) = sin(A + B) = 1 Rightarrow \
begin{cases}2A-B=0 \
A+B=frac{pi}{2} end{cases}Rightarrow begin{cases}A=frac{pi}{6} \ B=frac{pi}{3}end{cases} Rightarrow C=frac{pi}{2}.$$
Hence:
$$BC=ABsin frac{pi}{6}=2.$$
answered Nov 15 at 6:42
farruhota
17.7k2736
add a comment |
up vote
3
down vote
accepted
Since
$$cos(2A-B)le 1 text{and} sin(A + B)le 1;\
cos(2A-B) = sin(A + B) = 1 Rightarrow \
begin{cases}2A-B=0 \
A+B=frac{pi}{2} end{cases}Rightarrow begin{cases}A=frac{pi}{6} \ B=frac{pi}{3}end{cases} Rightarrow C=frac{pi}{2}.$$
Hence:
$$BC=ABsin frac{pi}{6}=2.$$
answered Nov 15 at 6:42
farruhota
17.7k2736
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Since
$$cos(2A-B)le 1 text{and} sin(A + B)le 1;\
cos(2A-B) = sin(A + B) = 1 Rightarrow \
begin{cases}2A-B=0 \
A+B=frac{pi}{2} end{cases}Rightarrow begin{cases}A=frac{pi}{6} \ B=frac{pi}{3}end{cases} Rightarrow C=frac{pi}{2}.$$
Hence:
$$BC=ABsin frac{pi}{6}=2.$$
answered Nov 15 at 6:42
farruhota
17.7k2736
Since
$$cos(2A-B)le 1 text{and} sin(A + B)le 1;\
cos(2A-B) = sin(A + B) = 1 Rightarrow \
begin{cases}2A-B=0 \
A+B=frac{pi}{2} end{cases}Rightarrow begin{cases}A=frac{pi}{6} \ B=frac{pi}{3}end{cases} Rightarrow C=frac{pi}{2}.$$
Hence:
$$BC=ABsin frac{pi}{6}=2.$$
answered Nov 15 at 6:42
farruhota
17.7k2736
answered Nov 15 at 6:42
farruhota
17.7k2736
answered Nov 15 at 6:42
farruhota
17.7k2736
answered Nov 15 at 6:42
farruhota
17.7k2736
17.7k2736
add a comment |
add a comment |
up vote
4
down vote
Hint:
For real $x,$ $$cos x,sin xle1$$
$$cos(2A-B) + sin(A + B) = 2impliescos(2A-B)=sin(A+B)=1$$
and $sin(A+B)=sin(pi-C)=sin C$
Can you take it from here?
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
add a comment |
up vote
4
down vote
Hint:
For real $x,$ $$cos x,sin xle1$$
$$cos(2A-B) + sin(A + B) = 2impliescos(2A-B)=sin(A+B)=1$$
and $sin(A+B)=sin(pi-C)=sin C$
Can you take it from here?
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
add a comment |
up vote
4
down vote
up vote
4
down vote
Hint:
For real $x,$ $$cos x,sin xle1$$
$$cos(2A-B) + sin(A + B) = 2impliescos(2A-B)=sin(A+B)=1$$
and $sin(A+B)=sin(pi-C)=sin C$
Can you take it from here?
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
Hint:
For real $x,$ $$cos x,sin xle1$$
$$cos(2A-B) + sin(A + B) = 2impliescos(2A-B)=sin(A+B)=1$$
and $sin(A+B)=sin(pi-C)=sin C$
Can you take it from here?
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
answered Nov 15 at 6:29
lab bhattacharjee
220k15154271
220k15154271
add a comment |
add a comment |
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