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- Feb 14, 2012

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\(\displaystyle \sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C\)

Find the measure of the angle $A$.

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- #1

- Feb 14, 2012

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\(\displaystyle \sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C\)

Find the measure of the angle $A$.

- Mar 31, 2013

- 1,309

\(\displaystyle \sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C\)

Find the measure of the angle $A$.

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

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- #3

Your technique is nice and elegant, but you have made a simple sign error...

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

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- #4

- Feb 14, 2012

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Well done,

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

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

- Mar 31, 2013

- 1,309

thanks markFL and anemoneWell done,kaliprasad...though you have made a minor mistake because $\ang A = 60 ^{\circ}$...

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

it should be

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

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- #6

The equation becomes:

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

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

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

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

Using Pythagorean identities, we have:

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

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

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

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

\(\displaystyle B+C=120^{\circ}\implies A=60^{\circ}\)

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- #7

- Feb 14, 2012

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Don't be sorry,thanks markFL and anemone

it should be

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

Bravo,

The equation becomes:

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

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

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

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

Using Pythagorean identities, we have:

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

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

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

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

\(\displaystyle B+C=120^{\circ}\implies A=60^{\circ}\)

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- #8

Now...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......

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- #9

- Feb 14, 2012

- 3,683

Very funny,Nowthere'sthe ultimate challenge!