Largest Angle of ABC

We're going to need the cosine rule for this problem.

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Now let's see what we have got. Basically we have nothing besides two equations that tell us the relation between the sides in a not-so-obvious way.

$a^2=a+2b+2c$

$-3=a+2b-2c$

When you look at the right hand sides of these equations, the first thing you would want to do is multiply them together. [Because you'd like to see something like this: $(p+q)(p-q)=p^2-q^2$ . There's no guarantee that it would you lead you somewhere, but it's worth a shot.]

So let's multiply them together.

We get, $-3a^2=(a+2b+2c)(a+2b-2c)=(a+2b)^2-(2c)^2=a^2+4ab+4b^2-4c^2$ .

After a little bit of rearranging, we have:

$4(a^2+b^2-c^2+ab)=0$ .

So, $a^2+b^2-c^2+ab=0$ $\cdots (1)$ .

Now it's time to use the cosine rule. From the cosine rule, $c^2=a^2+b^2-2ab\cos C$ . We can make this substitution in $(1)$ and all the squared terms cancel out.

And we're left with $ab(2\cos C +1)=0$ .

Since $ab$ isn't equal to $0$ , $2\cos C+1=0$ and that means $\cos C=-\frac{1}{2}$ .

Since $C$ is angle of a triangle, it is between $0$ and $180$ degrees. So the only solution for this equation is $C=120^{\circ}$ .

So, $C$ is an obtuse angle and angles $A$ and $B$ are acute. So $C$ is the largest angle of $\triangle ABC$ and it is equal to $\boxed{120}$ degrees.

By adding and subtracting the two given equations, we obtain expression for $b$ and $c$ respectively in terms of $a.$ We have $b=\frac{a^2-2a-3}{4}$ $c=\frac{a^2+3}{4}$ Clearly, $c>b.$ By the triangle inequality, we know that $a > c-b = \frac{3+a}{2},$ or $a>3.$ Now, since $\frac{a^2+3}{4}-a=\frac{(a-1)(a-3)}{4}$ is positive for all $a>3,$ $c>a$ as well.

Thus, the largest angle of $\triangle ABC$ is the angle opposite side $c,$ or $\angle C.$ By the law of cosines, $\cos \angle C=\frac{a^2+b^2-c^2}{2ab}=\frac{\frac{a^2+(a^2-2a-3)^2}{16}-\frac{(a^2+3)^2}{16}}{\frac{a^3-2a^2-3a}{2}}.$ This expression simplifies to $-\frac{1}{2}$ so $\angle C=120^\circ.$

We first show that for any $a > 3$ , there exists a $\triangle ABC$ satisfying the given conditions, and that for any such triangle, $c$ is always the longest side. Adding the two equations gives $a^2-3 = 2a + 4b$ , or $b = \frac{1}{4}(a^2-2a-3) = \frac{1}{4}(a-3)(a+1).$ It follows that $a, b > 0$ if and only if $a > 3$ . Taking the difference of the two equations gives $a^2+3 = 4c$ , or $c = \frac{1}{4}(a^2+3),$ which is always positive. So $c - b = \frac{1}{2}(a+3) > 0$ , and $c - a = \frac{1}{4}(a^2-4a+3) = \frac{1}{4}(a-3)(a-1) > 0$ , hence $c > b$ and $c > a$ . Next, we check the triangle inequalities: $c > b$ implies $a+c > b$ , and similarly, $c > a$ implies $b+c > a$ . Finally, $a+b-c = a - \frac{1}{2}(a+3) = \frac{1}{2}(a-3) > 0,$ so $a+b > c$ . Hence for any $a > 3$ , there is a triangle that satisfies the given conditions, whose longest side is $c$ , and therefore whose largest angle is $\angle ACB$ .

Next, we use the Law of Cosines to compute $\begin{aligned} \cos C &= \frac{a^2+b^2-c^2}{2ab} = \frac{a^2 - (c-b)(c+b)}{2ab} \\ &= \frac{a^2 - \frac{1}{4}(a+3)(a^2-a)}{\frac{1}{2}a(a-3)(a+1)} \\ &= \frac{4a-(a+3)(a-1)}{2(a-3)(a+1)} \\ &= -\frac{1}{2}, \end{aligned}$ and since $0 < \angle ACB < 180^\circ$ , it follows that $\angle ACB = 120^\circ$ .

Multiplying the two equations yields

-3a^2 = ((a+2b)+2c)((a+2b)-2c)

-3a^2 = (a+2b)^2 - (2c)^2

-3a^2 = a^2 + 4ab + 4b^2 - 4c^2

4a^2 + 4b^2 - 4c^2 = 4ab

So \frac{a^2 + b^2 - c^2}{2ab} = - \frac{1}{2}. But by the law of cosines, \frac{a^2 + b^2 - c^2}{2ab} = cos(C) (where C is the angle opposite the side with length c). So angle C must be equal to 120 ^ \circ. Furthermore, since the sum of the angles in a triangle is 180 ^ \circ, neither angle A nor angle B can exceed 60 ^ \circ. So the 120 ^ \circ angle must be the largest.

We know that $2b+2c=a^2-a$ and $2b-2c=-a-3$ . Given this, we can solve for $b$ and $c$ in terms of $a$ : $b=\frac{a^2-2a-3}{4}$ and $c=\frac{a^2+3}{4}$ . Since $a$ is the measure of a side, it must be positive. Thus, $c>b$ . For $a$ , $b$ and $c$ to be valid sides of a triangle, $a+b=\frac{a^2+2a-3}{4}$ should be greater than $c=\frac{a^2+3}{4}$ . Solving this inequality, we conclude that $a>3$ .

Now, we determine which is larger between $a$ and $c$ . Solving the inequality $a<\frac{a^2+3}{4}$ , we know that $a < c$ if $a < 1$ or $a > 3$ . Thus, $a<c$ and $c$ is the largest side.

Now we use cosine law to get the value of the angle opposite side $c$ : $\cos C=\displaystyle \frac{a^2+\left( \frac{a^2-2a-3}{4} \right)^2 - \left(\frac{a^2+3}{4}\right)^2}{2 \cdot a \cdot \frac{a^2-2a-3}{4}}$ Simplifying, $\cos C=-\frac{1}{2}$ . Thus, $m\angle C=120$ .

from $-3=a+2b-2c$ ,we know $b<c$ which could be useful in the later

and $a^2-2c=a+2b$ , and $-3+2c=a+2b$ , .

so $c= \frac{a^2+3}{4}$ , then use this in equation $a^2-2c=a+2b$

we get $b=\frac{1}{4}(a-3)(a+1)>0$ , so $a>3$

notice that when $a>3,c= \frac{a^2+3}{4}>a$

so $c$ is the biggest and $C$ is the largest angle.

$cos C=\frac{a^2+(b+c)(b-c)}{2ab}=\frac{a^2-\frac{a(a-1)(a+3)}{2*2}}{\frac{1}{2}a(a+1)(a-3)}$

simplify this and we get $cos C=-\frac{1}{2}$ so the answer is $120$ in degree

Since $-3=a+2b-2c$ , then $a=2c-2b-3$ . Substituting this into the other given equation and simplifying the resultant expression, we obtain $c^2-(2b+4)c+(b^2+3b+3)=0$ . Solving this as a quadratic equation in c, we get $c=b+2\pm\sqrt{b+1}$ . Substituting this back into the expression for a, we get $a=1\pm2\sqrt{b+1}$ . But since b is positive, $1-2\sqrt{b+1}<1-2\sqrt{1}<0$ . Since a is also positive, we can only have $a=1+2\sqrt{b+1}$ . Therefore, $c=b+2+\sqrt{b+1}$ . Now we calculate cos C using the cosine rule $c^2=a^2+b^2-2ab\cos C$ . Substituting the values of a and c (in terms of b) into the equation, the equation simplifies to $b(1+2\sqrt{b+1})=-2b(1+2\sqrt{b+1})\cos C$ . Since b and $1+2\sqrt{b+1}$ are both positive, then we can cancel these terms out on both sides to get $1=-2\cos C$ . This easily gives us $C=120^\circ$ . Since this is an obtuse angle, it must be the largest angle in the triangle. So the measure of the largest angle is $120^\circ$ .

$a^{2} = a + 2b + 2c$

$-3 = a + 2b - 2c$

So $a^{2} + 3 = 4c$

and $b = c - \frac{3}{2} - \frac{a}{2}$

First we'll determine the largest angle of the triangle. From our second equation above, and the fact that $a, b, c > 0$ , we know that $c > b$ . Furthermore, since the sum of the lengths of any two legs of a triangle must be greater than the length of the third leg, we know that $b + c > a$ .

From our first equation, then, it follows that

$a^{2} = a + 2b + 2c = a + 2(b + c) > a + 2a = 3a$

Thus $a > 3$ , and since $a^{2} + 3 = 4c$ , we know that $c > a$ . Therefore, $c$ is the longest side of the triangle, and $C$ is the largest angle.

Now we'll apply the law of cosines, relating sides and angles of a triangle:

$c^{2} = a^{2} + b^{2} - 2ab\cos C$

Substituting for a and b:

$c^{2} = 4c - 3 + (c - \frac{3}{2} - \frac{a}{2})^{2} - 2a(c - \frac{3}{2} - \frac{a}{2})\cos C$

$c^{2} = 4c - 3 + c^{2} - 3c + \frac{9}{4} - ac + \frac{3a}{2} + \frac{a^{2}}{4} - 2(ac - \frac{3a}{2} - \frac{a^{2}}{2})\cos C$

$c^{2} = 4c - 3 + c^{2} - 3c + \frac{9}{4} - ac + \frac{3a}{2} + c - \frac{3}{4} - 2(ac - \frac{3a}{2} - 2c + \frac{3}{2})\cos C$

$0 = (2c - \frac{3}{2} - ac + \frac{3a}{2}) + 2(2c - \frac{3}{2} - ac + \frac{3a}{2})\cos C$

Thus,

$-(2c - \frac{3}{2} - ac + \frac{3a}{2}) = (2c - \frac{3}{2} - ac + \frac{3a}{2})\cos C$

so

$\cos C = -\frac{1}{2}$

and

$\angle C = 120^\circ$

Let A B and C represent the angles BAC, CBA, and ACB respectively.

Observe that triangle inequality yields:

a^2 = a + 2b + 2c > 3a

a> 3

a^2 + 3 = 4c, and since a > 3, c > a

2c - 2b = a + 3

2c = 2b + a + 3

c > b

note that sinC/AB = sinA/BC = sin B/AB

and thus sin C > sin A and sin C > sin B

if A,B,C are all less than 90 degrees, C is the largest angle as the sine function is strictly increasing from 0 to 90 degrees, otherwise assume one of the angles A,B is >= 90 (WLOG assume its A), and we have:

let A = 90 + T where T >= 0

sinC > sin A = sin(90+T) = sin(90-T) = sin(B+C) However, B > 0 and thus 0 < C < B+C <= 90, implying sin(C) < sin(B+C), a contradiction.

Thus C is the largest angle. We now use cosine rule to solve for cosC which will give us angle C. We have:

a^2 = 4c - 3

a = rt(4c-3)

b = (2c - 3 - a)/2

Substituting these values for a and b into cosine rule, expanding and simple cancellation gives:

cosC = -1/2

which gives us the value of 120 degrees for angle C (actually, since this result is more than 90 degrees, we did not need to prove C was the largest earlier)

a^2 = a+2b+2c .........(1)

-3=a+2b-2c ...............(2)

add 1 and 2

then simplify to get

b= $\frac {a^2-3-2a}{4}$

subtract 1 and 2

c= $\frac {a^2 +3}{4}$

notice that b can be wirtten as

(a-1)^2/4 -1 so for b to be a positive number 'a' (the side a) should be greater than 3.

and also notice that a is always greater than b

also observe that c is greater than a for all a greater than 3 as (a-3)(a-1)>0 for all a greater than 3 (this fact is got from setting c>a and observing for what values c is greater than a)

hence the lengths of the sides in decreasing order is c>a>b

therefore the largest angle must be opposite the largest side.hence largest angle is angleC

using cosine rule

cos C = $\frac {a^2 + b^2 - c^2}{2ab}$

substituting equation 1 for b and equation 2 for c and simplifying we get

$\frac {16a^2 + (a^2 -3-2a)^2 - (a^2 + 3)^2}{8a(a^2-3-2a)}$

expand the terms and after cancelling them out we get cos C = -1/2

thus the angle C is 120 degrees

If we multiply $a^2=a+2b+2c$ and $-3=a+2b-2c$ ,we get $-3a^2=(a+2b)^2-4c^2$ , or equivalently, $c^2=a^2+b^2+ab$ . Comparing it with $c^2 =a^2+b^2-2ab\cos\theta$ ( $\theta$ is angle between sides $a$ and $b$ ), we get the desired answer of $120 ^ \circ$ .

(1) $a^2 = a + 2b + 2c$ (2) $-3 = a + 2b - 2c$

Adding (1) and (2), we get $a^2 - 3 = 2a +4b$ , so $b = \frac {a^2 - 2a - 3} {4}$ . Subtracting (2) from (1), we get $a^2 + 3 = 4c$ , so $c = \frac {a^2 + 3} {4}.$ . Thus, we can get c > b. Since the largest length corresponds to the largest angle, either C or A is the largest angle. Using the cosine rule, we get $\cos C = \frac {a^2 + b^2 - c^2} {2ab}$ . After substituting b and c in the expression in terms of a and simplifying, we get $\cos C = -\frac {1} {2}$ , so C = $120 ^ \circ$ and is the largest angle in the triangle.

$c=\frac{a^2+3}{4}$

$b=\frac{a^2-2a-3}{4}$

Using cosinus law:

$\angle ACB=\arccos -\frac{1}{2}$ (After Algebra Manipulation)

So, Maximum angle are $120^{\circ}$ ,

Measure angle maximum $\fbox{120}$

Since the clues, then c= 1/4 x a^2 + 3/4, b= 1/4 x a^2 - 1/2 x a - 3/4 = (1/2 x a - 1/2)^2 - 7/4 = 1/4 (a -1)^2 -7/4 a, b are the lengths of sides of the triangle, hence a, b>0 => 1/4 (a-1)^2 -7/4 >0 => (a-1)^2 >7 (a>0) => a>=4 Let a=4, then b= 5/4 and c= 19/4 and we have c is the largest side => angle C is the largest one. cos C = ( a^2 + b^2 - c^2) / (2 x a x b) = -1/2 => C=120

since we have from the given : a^{2}=4c-3 so we have to assume the lenght of c where it satisfies the two equations given luckily : c=21 satisfies so we get a=9 ,b=15 so by applying cosine law we get c=120

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We know that $a^2=a+2b+2$ and $-3=a+2b-2c$ . Adding these two equations we get $a^2-3=2a+4b$ or $a^2-2a-3=4b$ . Taking 4b=0(though we can't) and solving the quadratic in a , we get the roots 3and -1. Since the latter is not possible we come to know that a is NOT equal to 3, since b is not 0. Taking a=4, we get b=5/4. Putting this value in any of the 2 equations, we get c negative which is also not possible. Now let's take a=5, we get b=3 and c=7. Now we get a solution. Now comes the part of calculating the largest angle in the triangle from the three sides. First, we use the property that the angle opposite to the largest side is the largest. Thus we need to calculate angle opposite to c. Apply, the cosine rule , and we have the equation $c^2=a^2+b^2-2ab cosC$ . Substitute the values to have the final equation $49=25+9-30 cosC$ , cos C= -1/2 or angle C =120.

the given two equations a^2=a+2 b+ 2 c and −3=a+2 b−2 c

can be subtracted first

then we get the values of a,b,c c=7 b=3 a=5 then apply cosine formula we get -1/2 so the angle comes out to be 120

Subtract the second equation from the first to yield a^2 = 4c-3. Thus, a must be odd (because a^2=1(mod 4)). Trying small values of a, we see that (5,3,7) is a solution. This solution is unique because as a gets too large, b (from the first equation) would be greater than a + c, which is impossible from the Triangle Inequality. Using the Law of Cosines, we find that <C = 120.