Suppose, if here it was given that there are two sides known to us,i.e 10 and 12, then how many acute-angled triangles are possible to form ?? How do you deal with this question now

Angles can be classified into five groups, based on their measure in degrees.

Acute: angles with measure $< 90^\circ$

Right: angles with measure $= 90^\circ$

Obtuse: angles with measure $> 90^\circ$ and $< 180 ^ \circ$

Straight: angles with measure $= 180^ \circ$

Reflex: angles with measure $> 180^ \circ$> and $< 360 ^ \circ$

If $a, b,$ and $c$ are the $3$ sides of an acute triangle where $c$ is the longest side then $c^{2} < a^{2} + b^{2}$

The sides are $10, 12,$ and $'x'.$

Case$1:$ Among the $3$ sides $10, 12,$ and $x,$ for values of $x \leq 12,\; 12 $ is the longest side.

When $x \leq 12,$ let us find the number of values for $x$ that will satisfy the inequality $12^{2} < 10^{2} + x^{2}$

$\implies144 < 100 + x^{2}$

$\implies 44<x^{2}$

$\implies x^{2}>44$

The least integer value of $x$ that satisfies this inequality is $7.$

The inequality will hold true for $x = 7, 8, 9, 10, 11,$ and $12$. i.e., $6$ values.

Case$2:$ For values of $x > 12, x$ is the longest side.

Let us count the number of values of $x$ that will satisfy the inequality $x^{2} < 10^{2} + 12^{2}$
i.e., $x^{2} < 244$
$x = 13, 14,$ and $15$ satisfy the inequality. That is $3$ more values.

Hence, the values of $x$ for which $10, 12,$ and $x$ will form sides of an acute triangle are $x = 7, 8, 9, 10, 11, 12, 13, 14, 15.$