Trig identities: 4sinA+5cosB=5 and 5sinB+4cosA=6

[bobrossu]

Problem: In triangle ABC, where A and B are acute angles, 4sinA+5cosB=5 and 5sinB+4cosA=6. Find the radian measure of C.

Work: (4sinA+5cosB=5)^2+(5sinB+4cosA=6)^2 => use c^2=b^2+a^2 to get c^2 => simplify to 41+40(cosBsinA+cosAsinB)=61 (equal to c^2) => 40(sinA+B)=20 ....

Not sure what to do next to solve for C.

 

[ksdhart2]

Most of your work is good, except for a few things. One, the Pythagorean Theorem a² + b² = c² is about the sides of a triangle, not the angles. Two, the theorem only applies to right triangles, and you cannot guarantee that the arbitrary triangle described in the problem will be a right triangle.

Those corrections aside, once you've gotten to the last step of your work, you're almost there. If \(\displaystyle 40 \cdot sin(A+B) = 20\), then dividing through by 40 shows that \(\displaystyle sin(A + B) = \dfrac{1}{2}\). Keeping in mind that the three angles of a triangle must add up to 180 degrees (or pi radians), we can say that \(\displaystyle sin(\pi - C) = \dfrac{1}{2}\). Can you finish up from here?

 

[bobrossu]

Most of your work is good, except for a few things. One, the Pythagorean Theorem a² + b² = c² is about the sides of a triangle, not the angles. Two, the theorem only applies to right triangles, and you cannot guarantee that the arbitrary triangle described in the problem will be a right triangle.

Those corrections aside, once you've gotten to the last step of your work, you're almost there. If \(\displaystyle 40 \cdot sin(A+B) = 20\), then dividing through by 40 shows that \(\displaystyle sin(A + B) = \dfrac{1}{2}\). Keeping in mind that the three angles of a triangle must add up to 180 degrees (or pi radians), we can say that \(\displaystyle sin(\pi - C) = \dfrac{1}{2}\). Can you finish up from here?

Yes! Since sin(pi-C) = (1/2) then (pi-C) must equal to an angle (x), that sin(x)= (1/2). So (pi-C) = (pi/6) and C= (5pi/6)

But if I used the Pythagorean Theorem, wouldn't I have to square root something?

Also how would you have done it instead?
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[Deleted member 4993]

Yes! Since sin(pi-C) = (1/2) then (pi-C) must equal to an angle (x), that sin(x)= (1/2). So (pi-C) = (pi/6) and C= (5pi/6)

But if I used the Pythagorean Theorem, wouldn't I have to square root something?

Also how would you have done it instead?
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You CANNOT use Pythagorean theorem here.

To do that:

First you would have to prove that ABC is a right angled triangle.

 

[bobrossu]

You CANNOT use Pythagorean theorem here.

To do that:

First you would have to prove that ABC is a right angled triangle.

How would I prove it? Not sure what theorem(s) I could use.

Also If possible could someone show me an alternative way to solve it?

 

[ksdhart2]

Yes! Since sin(pi-C) = (1/2) then (pi-C) must equal to an angle (x), that sin(x)= (1/2). So (pi-C) = (pi/6) and C= (5pi/6)

But if I used the Pythagorean Theorem, wouldn't I have to square root something?

Also how would you have done it instead?
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Very good. But I'm a bit confused by your follow-up question. It appears to me you're asking about what you would do if the problem instead had boiled down to \(\displaystyle 12sin(\pi - C)=12\). If I'm correct in this interpretation, you would proceed in exactly the same way as before. Divide by 12 to isolate the sine, then take the inverse sine, such that \(\displaystyle \pi - C = sin^{-1}(1)\)

 

[bobrossu]

Very good. But I'm a bit confused by your follow-up question. It appears to me you're asking about what you would do if the problem instead had boiled down to \(\displaystyle 12sin(\pi - C)=12\). If I'm correct in this interpretation, you would proceed in exactly the same way as before. Divide by 12 to isolate the sine, then take the inverse sine, such that \(\displaystyle \pi - C = sin^{-1}(1)\)

That was a Typo, ignore it.

 

[Deleted member 4993]

Problem: In triangle ABC, where A and B are acute angles, 4sinA+5cosB=5 and 5sinB+4cosA=6. Find the radian measure of C.

Work: (4sinA+5cosB=5)^2+(5sinB+4cosA=6)^2 => use c^2=b^2+a^2 to get c^2 => simplify to 41+40(cosBsinA+cosAsinB)=61 (equal to c^2) => 40(sinA+B)=20 ....

Not sure what to do next to solve for C.

Clean-up
4sin(A)+5cos(B) = 5

16*sin²(A) + 25*cos²(B) + 20 sin(A)*cos(B) = 25...................................(1)

5sin(B) + 4cos(A) = 6

16*cos²(A) + 25*sin²(B) + 20 sin(B)*cos(A) = 36...................................(2)

(1) + (2)

16 + 25 + 20 * [sin(A)*cos(B) + sin(B)*cos(A)]= 61

sin(A + B) = 1/2 = sin (π/6) or sin (π - π/6) ............. continue....