MHB Find Angle A for Triangle: Sin2 B + Sin2 C = Sin B Sin C

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The discussion revolves around finding angle A in a triangle using the trigonometric equality sin²B + sin²C - sin²A = sinB sinC. Participants acknowledge a minor mistake in the calculations, confirming that angle A measures 60 degrees. There is a light-hearted exchange about the elegance of different methods used to solve the problem, with some participants expressing admiration for each other's approaches. The conversation encourages collaboration and sharing of more challenging problems in the future. Overall, the focus remains on the mathematical solution and the enjoyment of problem-solving.
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For the triangle with angles $A, B, C$, the following trigonometric equality holds.

$$\sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C$$

Find the measure of the angle $A$.
 
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anemone said:
For the triangle with angles $A, B, C$, the following trigonometric equality holds.

$$\sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C$$

Find the measure of the angle $A$.

using law of sin's sin A = ka, sin B= kb, sin C = kc

we get
b^2+c^2 - a^2 = bc

or a^2 = b^2 + c^2 + bc (1)

by law of cos

a^2 = b^2 + c^2 - 2bc cos A (2)

from (1) and (2)

2 cos A = - 1 or cos A = - 1/2 or A = 120 degrees
 
kaliprasad said:
using law of sin's sin A = ka, sin B= kb, sin C = kc

we get
b^2+c^2 - a^2 = bc

or a^2 = b^2 + c^2 + bc (1)

by law of cos

a^2 = b^2 + c^2 - 2bc cos A (2)

from (1) and (2)

2 cos A = - 1 or cos A = - 1/2 or A = 120 degrees

Your technique is nice and elegant, but you have made a simple sign error...:D
 
kaliprasad said:
using law of sin's sin A = ka, sin B= kb, sin C = kc

we get
b^2+c^2 - a^2 = bc

or a^2 = b^2 + c^2 + bc (1)

by law of cos

a^2 = b^2 + c^2 - 2bc cos A (2)

from (1) and (2)

2 cos A = - 1 or cos A = - 1/2 or A = 120 degrees

Well done, kaliprasad...though you have made a minor mistake because $\ang A = 60 ^{\circ}$...:p

Also, I think my silly method is not worth mentioning after reading to your method!:o
 
anemone said:
Well done, kaliprasad...though you have made a minor mistake because $\ang A = 60 ^{\circ}$...:p

Also, I think my silly method is not worth mentioning after reading to your method!:o

thanks markFL and anemone

it should be

a^2 = b^2 + c^2 - bc (1)

by law of cos

a^2 = b^2 + c^2 - 2bc cos A (2)

from (1) and (2)

2 cos A = 1 or cos A = 1/2 or A = 60 degrees
sorry for my mistake
 
My solution:

Since $$A=\pi-(B+C)$$ and $$\sin(\pi-\theta)=\sin(\theta)$$

The equation becomes:

$$\sin^2(B)+\sin^2(C)-\sin^2(B+C)=\sin(B)\sin(C)$$

Using the angle-sum identity for sine, we may write:

$$\sin^2(B)+\sin^2(C)-\sin^2(B)\cos^2(C)-2\sin(B)\sin(C)\cos(B)\cos(C)-\sin^2(C)\cos^2(B)=\sin(B)\sin(C)$$

$$\sin^2(B)\left(1-\cos^2(C) \right)+sin^2(C)\left(1-\cos^2(B) \right)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$

Using Pythagorean identities, we have:

$$2\sin^2(B)\sin^2(C)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$

Assuming the triangle is not degenerate, i.e., $$\sin(B)\sin(C)\ne0$$ we have:

$$2\sin(B)\sin(C)=1+2\cos(B)\cos(C)$$

$$-\frac{1}{2}=\cos(B)\cos(C)-\sin(B)\sin(C)=\cos(B+C)$$

$$B+C=120^{\circ}\implies A=60^{\circ}$$
 
kaliprasad said:
thanks markFL and anemone

it should be

a^2 = b^2 + c^2 - bc (1)

by law of cos

a^2 = b^2 + c^2 - 2bc cos A (2)

from (1) and (2)

2 cos A = 1 or cos A = 1/2 or A = 60 degrees
sorry for my mistake

Don't be sorry, kali! And if you would like, you can make it up by joining me to post many interesting and challenging problems for the folks here...hehehe...(Emo)

MarkFL said:
My solution:

Since $$A=\pi-(B+C)$$ and $$\sin(\pi-\theta)=\sin(\theta)$$

The equation becomes:

$$\sin^2(B)+\sin^2(C)-\sin^2(B+C)=\sin(B)\sin(C)$$

Using the angle-sum identity for sine, we may write:

$$\sin^2(B)+\sin^2(C)-\sin^2(B)\cos^2(C)-2\sin(B)\sin(C)\cos(B)\cos(C)-\sin^2(C)\cos^2(B)=\sin(B)\sin(C)$$

$$\sin^2(B)\left(1-\cos^2(C) \right)+sin^2(C)\left(1-\cos^2(B) \right)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$

Using Pythagorean identities, we have:

$$2\sin^2(B)\sin^2(C)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$

Assuming the triangle is not degenerate, i.e., $$\sin(B)\sin(C)\ne0$$ we have:

$$2\sin(B)\sin(C)=1+2\cos(B)\cos(C)$$

$$-\frac{1}{2}=\cos(B)\cos(C)-\sin(B)\sin(C)=\cos(B+C)$$

$$B+C=120^{\circ}\implies A=60^{\circ}$$

Bravo, MarkFL! And I like your method as well! Just so you know, my method is so much longer and more tedious than yours! (Angry)
 
anemone said:
...And if you would like, you can make it up by joining me to post many interesting and challenging problems for the folks here...hehehe...(Emo)...

(Rofl) Now there's the ultimate challenge! (Wave) (Nerd)
 
MarkFL said:
(Rofl) Now there's the ultimate challenge! (Wave) (Nerd)

Very funny, MarkFL!(Rofl)
 
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