In ΔABC, right angled at B. AB = 24 cm, BC = 7 cm. Determine:
(i) sin A cos A,
(ii) sin C, cos C.

Let AB = 24 cm
BC = 7cm Using Pythagoras theorem, we have
AC² = AB² + BC²
= (24 cm)²+ (7 cm)²
= 576 cm² + 49 cm²
= 625 cm²
So, AC = 25 cm

Now,

It is given that:
15 cot A = 8
              

Let
AB = 8K, BC = 15K
Using Pythagoras theorem, we have
AC² = AB² + BC² = (8K)² + (15K)²
= 64K² + 225K²
= 289K²
So,
AC = 17K
Now,
AC = 17K

Let us draw a right angle triangle, right angled at B.
We know that:
  
Let AB = 12 K,   AC = 13K
where K is a positive number.

Using Pythagoras theorem, we have
AC² = AB² + BC²
⇒(13 K)² = (12 K)² + BC²
⇒169 K² = 144 K² + BC²
⇒BC² = 169 K²-144 K²
BC² = 25 K²  BC = 5 K.
Now, 


and 

Let
PQ = 12K
and
PR = 13K
Using Pythagoras theorem, we have
PR² = PQ² + QR²
⇒(13K)² = (12K)² + QR²
⇒169K² = 144K² + QR²
⇒QR² = 169K² - 144K²
⇒QR² = 25K²
So, QR = 5K

Now, 

Therefore,