Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?

share|improve this question
    
Yesterday, itself i told you that whenever you see this type of things please apply the sine rule. –  anonymous Nov 17 '10 at 7:53

3 Answers 3

up vote 8 down vote accepted

This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.)

Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$.

Expand $\sin(A+B)$ and do the manipulations to get

$2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$

which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\cos(A)\cos(B) = \sin(A)\sin(B) \Rightarrow \cos(A+B) = 0 \Rightarrow \cos(C) = 0$.

Hence either $A = \pi/2$ or $B = \pi/2$ or $C = \pi/2$.

So the triangle is a right-angled triangle.

share|improve this answer

In general, in $\triangle{ABC}$

If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.

If $\sin^2 A + \sin^2 B + \sin^2 C = 2$ then $\triangle{ABC}$ is right angled.

If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.

Assume $A \le B \lt \pi/2$ and $ A \le B \le C$.

Basically, if $k = \sin^2 A + \sin^2 B + \sin^2 C$

then we have that

$3-2k = \cos 2A + \cos 2B + \cos (2A+2B)$

i.e

$3-2k = 2\cos(A+B)\cos(A-B) + 2\cos^2(A+B) -1 $

i.e

$4-2k = 4\cos(A+B)\cos A\cos B$

So if $k > 2$, then $\cos(A+B) \lt 0$ hence acute.

$k = 2$, then $\cos(A+B) = 0$ hence right triangled.

$k < 2$, then $\cos(A+B) \gt 0$, hence obtuse.

In fact, we can go further and show that the maximum possible value of $k$ is $k = \frac{9}{4}$ which corresponds to $\triangle{ABC}$ being equilateral, as follows:

$4\sin^2 A + 4\sin^2 B + 4\sin^2 C = 9 + \delta$

i.e.

$(2 - 2\cos2A) + (2-2\cos 2B) + 4(1- \cos^2 (A+B)) = 9 + \delta$

i.e. $1 + 2\cos2A + 2\cos 2B + 4\cos^2(A+B) = -\delta$

i.e.

$1 + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$

i.e.

$\sin^2(A-B) + \cos^2(A-B) + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$

i.e.

$\sin^2(A-B) + (\cos (A-B) + 2\cos(A+B))^2 = -\delta$.

Hence $\delta \le 0$ and so $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$

The case $\delta = 0$ gives us $\sin(A-B) = 0$ and $\cos(A+B) = \frac{-1}{2}$.

Hence $A=B=C$.

Thus the max value of $\sin^2 A + \sin^2 B + \sin^2 C$ is $\frac{9}{4}$ and is achieved when $A=B=C$.

share|improve this answer

$$\sin^2A+\sin^2B+\sin^2C$$ $$=1-(\cos^2A-\sin^2B)+1-\cos^2C$$ $$=2-\cos(A+B)\cos(A-B)-\cos C\cdot\cos C$$ $$=2-\cos(\pi-C)\cos(A-B)-\cos\{\pi-(A+B)\}\cdot\cos C$$ $$=2+\cos C\cos(A-B)+\cos(A+B)\cdot\cos C\text{ as }\cos(\pi-x)=-\cos x$$ $$=2+\cos C\{\cos(A-B)+\cos(A+B)\}$$ $$=2+2\cos A\cos B\cos C$$

$(1)$ If $2+2\cos A\cos B\cos C=2, \cos A\cos B\cos C=0$

$\implies $ at least one of $\cos A,\cos B,\cos C$ is $0$ which needs the respective angles $=\frac\pi2$

But we can have at most one angle $\ge \frac\pi2$

So, here we shall have exactly one angle $=\frac\pi2$

$(2)$ If $2+2\cos A\cos B\cos C>2, \cos A\cos B\cos C>0$

Either all of $\cos A,\cos B,\cos C$ must be $>0\implies$ all the angles are acute

or exactly two cosine ratios $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible for a triangle

$(3)$ If $2+2\cos A\cos B\cos C<2, \cos A\cos B\cos C<0$

Either all the ratios $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible fro a triangle

or exactly one of the cosine ratios is $<0\implies $ the respective angle $> \frac\pi2,$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.