Mistakes give rise to Problems- 17

Here, we have to find the number of ordered triples of integers $(a,b,\theta)$ such that $0\leq a,b\leq 10$ , $0\leq \theta \leq 180$ and $a\times b \times \left|\cos\theta \right| = a\times b \times \left|\sin\theta \right|$

This is simply $a\times b\times \bigl(\left|\sin\theta \right|-\left|\cos\theta \right| \bigr)=0$

Now this is true at $a=0$ or $b=0$ or $\theta = 45^\circ \text{ or } 135^\circ$

$\bullet \quad \quad a=0$

$(a,b,\theta )=(0,b,\theta)$ , $b$ has $11$ choices (0 to 10) and $\theta$ has $181$ choices, giving $181\times 11$ triples.

$\bullet \quad \quad b=0$

$(a,b,\theta)=(a,0,\theta)$ again same thing, but now $a$ has $10$ choices and $\theta$ has 181. (Because $(0,0,\theta)$ have been counted in previous case.) This gives $10\times 181$ triples.

$\bullet \quad \quad \theta = 45^\circ$ or $135^\circ$

$(a,b,\theta)$ , we have $10$ choices for $a$ and $10$ for $b$ (they are now non-zero), giving $10\times 10\times 2=200$ triples.

Total triples are $181\times 11 +181 \times 10 +200 = 181\times 21 +200 = \boxed{4001}$

The solution is to find the number $N$ of ordered integral triplets $(a, b, \theta)$ that satisfy $|ab \cos{\theta}|=|ab \sin{\theta} |$ .

There are two cases.

Case 1 : when $|ab \cos{\theta}|=|ab \sin{\theta} | = 0$ ; this happens when $a=0$ or $b=0$ . There are a total of 21 such cases as follows:

$(a,b) = (0,0), (0,1), (0,2), (0,3), (0,4), (0,5), (0,6), (0,7), (0,8), (0,9), (0,10),$

$\quad \quad \quad \space (1,0), (2,0), (3,0), (4,0), (5,0), (6,0), (7,0), (8,0), (9,0), (10,0)$

For each case of $a,b=0$ , $\theta$ can be $0^\circ, 1^\circ, 2^\circ ...$ or $180^\circ$ , a total of $181$ cases.

Therefore the number of ordered triplets in Case 1:

$N_1 = 21 \times 181 = 3801$

Case 2 : when $|ab \cos{\theta}|=|ab \sin{\theta} | \ne 0$ ; then:

$|ab \cos{\theta}|=|ab \sin{\theta} | =|\sin{\theta} | \quad \Rightarrow |\tan{\theta} | = | 1 | \quad \Rightarrow \tan{\theta} = \pm 1 \quad \Rightarrow \theta = 45^o, 135^o$

It can be seen that, in this case, $a$ can be any of $1, 2, 3 ... 10$ for every case of $b = 1, 2, 3 ... 10$ .

Therefore the number of ordered triplets in Case 2:

$N_2 = 10 \times 10 \times 2 = 200$

Therefore, $N = N_1 + N_2 = 3801 + 200 = \boxed{4001}$