If the sides of the Triangle are $a$, $b$ and $sqrt(a^2+ab+b^2)$, then find the value of the greatest Angle.

Clash Royale CLAN TAG#URR8PPP

up vote
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down vote

favorite

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If the sides of the Triangle are $a$, $b$ and $sqrt(a^2+ab+b^2)$, then find the value of the greatest Angle.

Answer Given in my notebook - $frac2pi3$ radians.

The Question is from the Chapter Solutions of Triangles and I'm aware of Sine and Cosine Rule of the triangles.

---

My try.

In any case, $a$ or $b$, no matter which side is greater, the 3rd side (Let's denote it by c) $ sqrt(a^2+ab+b^2) $ will have the greatest angle (i.e. greatest angle will be opposite to this side).

Now, let's take Opposite Angles of sides a, b and c as $A$, $B$, and $C$ respectively.
Then the only way to find the angle formed at angle C of the triangle can be easily calculated using Cosine Rule (if all the angles are given, which in this case is true), so putting the formula I get -

$$cos C = fraca^2+b^2-c^22ab = sqrt(a^2+ab+b^2)$$

Now, in the following case, I've to get the Value of Cosine function as $120^o or frac2pi3$ which I'm not sure how to do.

Can anyone help me with that?

Thanks :)

trigonometry triangle problem-solving

share|cite|improve this question

edited Jul 25 at 15:09

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You should have a $2$ in the denominator is your formula for the cosine.
  – saulspatz
  Jul 25 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Thanks, but still it's difficult for me to solve
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  rather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2abcos C implies cos C = -frac 12$
  – Doug M
  Jul 25 at 15:27
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

If the sides of the Triangle are $a$, $b$ and $sqrt(a^2+ab+b^2)$, then find the value of the greatest Angle.

Answer Given in my notebook - $frac2pi3$ radians.

The Question is from the Chapter Solutions of Triangles and I'm aware of Sine and Cosine Rule of the triangles.

---

My try.

In any case, $a$ or $b$, no matter which side is greater, the 3rd side (Let's denote it by c) $ sqrt(a^2+ab+b^2) $ will have the greatest angle (i.e. greatest angle will be opposite to this side).

Now, let's take Opposite Angles of sides a, b and c as $A$, $B$, and $C$ respectively.
Then the only way to find the angle formed at angle C of the triangle can be easily calculated using Cosine Rule (if all the angles are given, which in this case is true), so putting the formula I get -

$$cos C = fraca^2+b^2-c^22ab = sqrt(a^2+ab+b^2)$$

Now, in the following case, I've to get the Value of Cosine function as $120^o or frac2pi3$ which I'm not sure how to do.

Can anyone help me with that?

Thanks :)

trigonometry triangle problem-solving

share|cite|improve this question

edited Jul 25 at 15:09

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You should have a $2$ in the denominator is your formula for the cosine.
  – saulspatz
  Jul 25 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Thanks, but still it's difficult for me to solve
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  rather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2abcos C implies cos C = -frac 12$
  – Doug M
  Jul 25 at 15:27
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

If the sides of the Triangle are $a$, $b$ and $sqrt(a^2+ab+b^2)$, then find the value of the greatest Angle.

Answer Given in my notebook - $frac2pi3$ radians.

The Question is from the Chapter Solutions of Triangles and I'm aware of Sine and Cosine Rule of the triangles.

---

My try.

In any case, $a$ or $b$, no matter which side is greater, the 3rd side (Let's denote it by c) $ sqrt(a^2+ab+b^2) $ will have the greatest angle (i.e. greatest angle will be opposite to this side).

Now, let's take Opposite Angles of sides a, b and c as $A$, $B$, and $C$ respectively.
Then the only way to find the angle formed at angle C of the triangle can be easily calculated using Cosine Rule (if all the angles are given, which in this case is true), so putting the formula I get -

$$cos C = fraca^2+b^2-c^22ab = sqrt(a^2+ab+b^2)$$

Now, in the following case, I've to get the Value of Cosine function as $120^o or frac2pi3$ which I'm not sure how to do.

Can anyone help me with that?

Thanks :)

trigonometry triangle problem-solving

share|cite|improve this question

edited Jul 25 at 15:09

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

If the sides of the Triangle are $a$, $b$ and $sqrt(a^2+ab+b^2)$, then find the value of the greatest Angle.

Answer Given in my notebook - $frac2pi3$ radians.

The Question is from the Chapter Solutions of Triangles and I'm aware of Sine and Cosine Rule of the triangles.

---

My try.

In any case, $a$ or $b$, no matter which side is greater, the 3rd side (Let's denote it by c) $ sqrt(a^2+ab+b^2) $ will have the greatest angle (i.e. greatest angle will be opposite to this side).

Now, let's take Opposite Angles of sides a, b and c as $A$, $B$, and $C$ respectively.
Then the only way to find the angle formed at angle C of the triangle can be easily calculated using Cosine Rule (if all the angles are given, which in this case is true), so putting the formula I get -

$$cos C = fraca^2+b^2-c^22ab = sqrt(a^2+ab+b^2)$$

Now, in the following case, I've to get the Value of Cosine function as $120^o or frac2pi3$ which I'm not sure how to do.

Can anyone help me with that?

Thanks :)

trigonometry triangle problem-solving

share|cite|improve this question

edited Jul 25 at 15:09

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 15:09

edited Jul 25 at 15:09

edited Jul 25 at 15:09

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

asked Jul 25 at 15:05

Abhas Kumar Sinha

10011

10011

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You should have a $2$ in the denominator is your formula for the cosine.
  – saulspatz
  Jul 25 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Thanks, but still it's difficult for me to solve
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  rather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2abcos C implies cos C = -frac 12$
  – Doug M
  Jul 25 at 15:27
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You should have a $2$ in the denominator is your formula for the cosine.
  – saulspatz
  Jul 25 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Thanks, but still it's difficult for me to solve
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  rather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2abcos C implies cos C = -frac 12$
  – Doug M
  Jul 25 at 15:27
  
  

  
  
  
  

1

1

You should have a $2$ in the denominator is your formula for the cosine.
– saulspatz
Jul 25 at 15:08

You should have a $2$ in the denominator is your formula for the cosine.
– saulspatz
Jul 25 at 15:08

@saulspatz Thanks, but still it's difficult for me to solve
– Abhas Kumar Sinha
Jul 25 at 15:10

@saulspatz Thanks, but still it's difficult for me to solve
– Abhas Kumar Sinha
Jul 25 at 15:10

1

1

rather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2abcos C implies cos C = -frac 12$
– Doug M
Jul 25 at 15:27

rather that the formual abouve, I think the law of cosines directly is more intiufitve. $c^2 = a^2 + b^2 - 2abcos C implies cos C = -frac 12$
– Doug M
Jul 25 at 15:27

add a comment | 

2 Answers
2

active

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up vote
2
down vote

$$cos C=fraca^2+b^2-c^22abtag1$$
as you nearly say, but $c$ is $sqrta^2+ab+b^2$. Substituting that
into $(1)$ should give you the $C$ that you want.

share|cite|improve this answer

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I'm unable to do, How can you solve such equations?
  – Abhas Kumar Sinha
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No need to solve: substitute!
  – Lord Shark the Unknown
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Substitute what in where?
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Substitute $sqrta^2+ab+b^2$ for $c$
  – saulspatz
  Jul 25 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
  – Abhas Kumar Sinha
  Jul 25 at 15:11
  

  
  
  
  

 | 
show 3 more comments

up vote
0
down vote

In Triangle ABC,

the side $c=sqrta^2+ab+b^2$ ----------------------(1)

and by cosine Rule, -

$cos C = fraca^2+b^2-c^22ab$ ------------------------------(2)

So, substituting (1) in (2) gives,

$$cos C = fraca^2+b^2-(a^2+ab+b^2)2ab$$
$$Rightarrow cos C = frac-ab2ab$$
$$iff C = 2npi pm frac2pi3$$ (for $n$ as any integer)

The rule of the Sum angle Property only stands correct only when $n=1$, thus $120^o$ or $frac2pi3$ is the reqd. answer

share|cite|improve this answer

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

$$cos C=fraca^2+b^2-c^22abtag1$$
as you nearly say, but $c$ is $sqrta^2+ab+b^2$. Substituting that
into $(1)$ should give you the $C$ that you want.

share|cite|improve this answer

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I'm unable to do, How can you solve such equations?
  – Abhas Kumar Sinha
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No need to solve: substitute!
  – Lord Shark the Unknown
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Substitute what in where?
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Substitute $sqrta^2+ab+b^2$ for $c$
  – saulspatz
  Jul 25 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
  – Abhas Kumar Sinha
  Jul 25 at 15:11
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

$$cos C=fraca^2+b^2-c^22abtag1$$
as you nearly say, but $c$ is $sqrta^2+ab+b^2$. Substituting that
into $(1)$ should give you the $C$ that you want.

share|cite|improve this answer

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I'm unable to do, How can you solve such equations?
  – Abhas Kumar Sinha
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No need to solve: substitute!
  – Lord Shark the Unknown
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Substitute what in where?
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Substitute $sqrta^2+ab+b^2$ for $c$
  – saulspatz
  Jul 25 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
  – Abhas Kumar Sinha
  Jul 25 at 15:11
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

up vote
2
down vote

$$cos C=fraca^2+b^2-c^22abtag1$$
as you nearly say, but $c$ is $sqrta^2+ab+b^2$. Substituting that
into $(1)$ should give you the $C$ that you want.

share|cite|improve this answer

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

$$cos C=fraca^2+b^2-c^22abtag1$$
as you nearly say, but $c$ is $sqrta^2+ab+b^2$. Substituting that
into $(1)$ should give you the $C$ that you want.

share|cite|improve this answer

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

answered Jul 25 at 15:08

Lord Shark the Unknown

84.9k950111

84.9k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I'm unable to do, How can you solve such equations?
  – Abhas Kumar Sinha
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No need to solve: substitute!
  – Lord Shark the Unknown
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Substitute what in where?
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Substitute $sqrta^2+ab+b^2$ for $c$
  – saulspatz
  Jul 25 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
  – Abhas Kumar Sinha
  Jul 25 at 15:11
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I'm unable to do, How can you solve such equations?
  – Abhas Kumar Sinha
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  No need to solve: substitute!
  – Lord Shark the Unknown
  Jul 25 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Substitute what in where?
  – Abhas Kumar Sinha
  Jul 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Substitute $sqrta^2+ab+b^2$ for $c$
  – saulspatz
  Jul 25 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
  – Abhas Kumar Sinha
  Jul 25 at 15:11
  

  
  
  
  

That's what I'm unable to do, How can you solve such equations?
– Abhas Kumar Sinha
Jul 25 at 15:09

That's what I'm unable to do, How can you solve such equations?
– Abhas Kumar Sinha
Jul 25 at 15:09

1

1

No need to solve: substitute!
– Lord Shark the Unknown
Jul 25 at 15:09

No need to solve: substitute!
– Lord Shark the Unknown
Jul 25 at 15:09

Substitute what in where?
– Abhas Kumar Sinha
Jul 25 at 15:10

Substitute what in where?
– Abhas Kumar Sinha
Jul 25 at 15:10

1

1

Substitute $sqrta^2+ab+b^2$ for $c$
– saulspatz
Jul 25 at 15:11

Substitute $sqrta^2+ab+b^2$ for $c$
– saulspatz
Jul 25 at 15:11

oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
– Abhas Kumar Sinha
Jul 25 at 15:11

oh okay, I followed it, by mistake, I substituted that size side c, in angle C of cos..... My bad XD :)
– Abhas Kumar Sinha
Jul 25 at 15:11

 | 
show 3 more comments

up vote
0
down vote

In Triangle ABC,

the side $c=sqrta^2+ab+b^2$ ----------------------(1)

and by cosine Rule, -

$cos C = fraca^2+b^2-c^22ab$ ------------------------------(2)

So, substituting (1) in (2) gives,

$$cos C = fraca^2+b^2-(a^2+ab+b^2)2ab$$
$$Rightarrow cos C = frac-ab2ab$$
$$iff C = 2npi pm frac2pi3$$ (for $n$ as any integer)

The rule of the Sum angle Property only stands correct only when $n=1$, thus $120^o$ or $frac2pi3$ is the reqd. answer

share|cite|improve this answer

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

add a comment | 

up vote
0
down vote

In Triangle ABC,

the side $c=sqrta^2+ab+b^2$ ----------------------(1)

and by cosine Rule, -

$cos C = fraca^2+b^2-c^22ab$ ------------------------------(2)

So, substituting (1) in (2) gives,

$$cos C = fraca^2+b^2-(a^2+ab+b^2)2ab$$
$$Rightarrow cos C = frac-ab2ab$$
$$iff C = 2npi pm frac2pi3$$ (for $n$ as any integer)

The rule of the Sum angle Property only stands correct only when $n=1$, thus $120^o$ or $frac2pi3$ is the reqd. answer

share|cite|improve this answer

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

add a comment | 

up vote
0
down vote

up vote
0
down vote

In Triangle ABC,

the side $c=sqrta^2+ab+b^2$ ----------------------(1)

and by cosine Rule, -

$cos C = fraca^2+b^2-c^22ab$ ------------------------------(2)

So, substituting (1) in (2) gives,

$$cos C = fraca^2+b^2-(a^2+ab+b^2)2ab$$
$$Rightarrow cos C = frac-ab2ab$$
$$iff C = 2npi pm frac2pi3$$ (for $n$ as any integer)

The rule of the Sum angle Property only stands correct only when $n=1$, thus $120^o$ or $frac2pi3$ is the reqd. answer

share|cite|improve this answer

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

In Triangle ABC,

the side $c=sqrta^2+ab+b^2$ ----------------------(1)

and by cosine Rule, -

$cos C = fraca^2+b^2-c^22ab$ ------------------------------(2)

So, substituting (1) in (2) gives,

$$cos C = fraca^2+b^2-(a^2+ab+b^2)2ab$$
$$Rightarrow cos C = frac-ab2ab$$
$$iff C = 2npi pm frac2pi3$$ (for $n$ as any integer)

The rule of the Sum angle Property only stands correct only when $n=1$, thus $120^o$ or $frac2pi3$ is the reqd. answer

share|cite|improve this answer

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

answered Jul 25 at 15:26

Abhas Kumar Sinha

10011

10011

add a comment | 

add a comment | 

 

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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

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Post as a guest

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Name Email

Name

Email

Name Email

Name

Email