Question: Is there any short cut method to find the value of $a$ used in the expressions:

> 
> * $\sin x = {a^2+24^2-15^2\over2\times24a}$
> * $\cos x={a^2+7^2-15^2\over2\times7a}$
> 
> 
>

Answer: 
Hint: Since $$|\sin(x)|=\left|\frac{a^2+24^2-15^2}{28a}\right|\le 1$$ we get
$$a^2+351\le 48|a|$$
Can you solve this inequality? The solution is given by $$-39\le a\le -9$$ or $$9\le a\le 39$$
By the same idea we get $$|a^2-176|\le 22|a|$$ with the solution $$-22\le a\le -8$$ or $$8\le a\le 22$$
In the first case, $24,15,a$ are three sides of a triangle if we set $\dfrac\pi2-x=y$

Using <https://www.varsitytutors.com/hotmath/hotmath_help/topics/triangle-inequality-theorem>

$24+15>a$

$24+a>15$

$a+15>24$

$\implies9<a<39$

Similarly on the second, $a,7,15$ are the sides of a triangle

$\implies 8<a<22$

But $$\cos^2x+\sin^2x=1$$ should not be tough to handle