Question from trigonometry -- prove that the largest angle is greater than 120°

[prashant singh]

In a triangle whose sides are 3,4 and root 38 metres respectively, prove that the largest angle is greater than 120°

• My answer: Here angle C > B...(1) and C > A ...(2) adding (1) and (2) we get 2c > A+B, taking sine both side , sin2C = sin(A+B) = sin(C), therefore cosc > (1/2) therefore angle C> 60°
• I am not getting C> 120°, what is the problem please help

 

[Hiranya Pasan]

(180-60)=120 is also a solution, now you should think how to prove that particular solution

 

[mfig]

Why not use the law of cosines? Find the angle between the short sides.

$38 = 3^2 + 4^2 - 2(3)(4)cos(θ)$

Now solve for θ.

 

[SteamKing]

Why not use the law of cosines? Find the angle between the short sides.

$38 = 3^2 + 4^2 - 2(3)(4)cos(θ)$

Now solve for θ.

There's a typo in the equation above. The 38 should also be squared.

 

[SammyS]

There's a typo in the equation above. The 38 should also be squared.

No. It's $\ \sqrt{38}\$ which is being squared.

 

[SteamKing]

No. It's $\ \sqrt{38}\$ which is being squared.

Missed the root in the OP.

 

[haruspex]

taking sine both side , sin2C = sin(A+B) = sin(C),

I guess you meant sin 2C > sin(A+B), but that does not follow. Once the angle exceeds 90 degrees the sine function is decreasing. Indeed, sin(2C) would be < sin(A+B) here.

cosc > (1/2) therefore angle C> 60°

That's backwards. If sin(2C) were > sin(A+B), it would follow that C is less than 60.

@mfig's method is fine, but you do not need to solve for theta. Just show that the longest side is too long for a 120 angle.