Welcome to our community

Be a part of something great, join today!

Find the angle A

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
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 $A$.
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
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

Administrator
Staff member
Feb 24, 2012
13,775
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
 
  • Thread starter
  • Admin
  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
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!:eek:
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
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!:eek:
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

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

\(\displaystyle B+C=120^{\circ}\implies A=60^{\circ}\)
 
  • Thread starter
  • Admin
  • #7

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
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)

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! (Angry)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...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)
 
  • Thread starter
  • Admin
  • #9

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683