Find the angle A

anemone

MHB POTW Director
Staff member
For the triangle with angles $A, B, C$, the following trigonometric equality holds.

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

Find the measure of the angle $A$.

Well-known member
For the triangle with angles $A, B, C$, the following trigonometric equality holds.

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

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

MarkFL

Staff member
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... anemone

MHB POTW Director
Staff member
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 $\ang A = 60 ^{\circ}$... Also, I think my silly method is not worth mentioning after reading to your method! Well-known member
Well 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! 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

MarkFL

Staff member
My solution:

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

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:

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

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

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

anemone

MHB POTW Director
Staff member
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... My solution:

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

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:

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

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

$$\displaystyle B+C=120^{\circ}\implies A=60^{\circ}$$
Bravo, MarkFL! And I like your method as well! Just so you know, my method is so much longer and more tedious than yours! MarkFL

...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... ... Now there's the ultimate challenge!   Now there's the ultimate challenge!  Very funny, MarkFL!!! 