since in a triangle each side must be shorter than the sum of the other 2 sides (otherwise the end points cannot connect, and there is no triangle), the necessary inequality condition must be

side < 1 + 2 = 3

so,

side < 3

for a lower limit let's go through the cases

1 < 2 + side (is always true)

2 < 1 + side

1 < side (side must be larger than 1)

and again

side < 1 + 2 = 3

side < 3

so the full restriction for the third side is

1 < side < 3

7 0

2 years ago

(x-4)2 + (y + 3)2 =

White raven [17]

Answer:

the answer is 2x+2y-2

then all you need to do is write in standard form.

7 0

4 years ago

If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn

mash [69]

Let $a=\tan A$ and $b=\sin A$ . Then

$m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab$

$\implies m^2-n^2=4\tan A\sin A$

and

$mn=(a+b)(a-b)=a^2-b^2$

$\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}$

The expression under the square root can be rewritten as

$\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)$

Recall that

$\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A$

so that

$\tan^2A-\sin^2A=\sin^2A\tan^2A$

and assuming $\sin A>0$ and $\tan A>0$ , we end up with

$4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A$

so that

$m^2-n^2=4\sqrt{mn}$

as required.

5 0

3 years ago