We shall use the formula for sum of sin of angles in AP
We have
$ \sum_{k=1}^{n}\sin\,kx = \dfrac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin \frac{x}{2}}$
Put n = 35 and $ x = 5^\circ$
To get
$ \sum_{k=1}^{35}\sin\,5k^\circ = \frac{\sin\dfrac{175}{2}^\circ\sin\,90^\circ}{\sin \dfrac{5}{2}^\circ}$
$ = \dfrac{\sin\dfrac{175}{2}^\circ}{\cos \dfrac{175}{2}^\circ}=\tan \dfrac{175}{2}^\circ$
From given condition $a=175,b=2,\dfrac{a}{b} = 87.5 <90$
So $a+b = 177$
We are given $a^{b^c} = a^{bc}$
Or $a^{b^c-bc} = 1$
This has got 2 sets of solutions
1) a = 1 and b and c can be any positive integer
2) a = any positive integer other than 1 and
$b^c - bc = 0$
Or $b(b^{c-1}) = c$
As b is not zero so we have $ b^{c-1} = c $
if c = 1 we get b any number
Or $b= \sqrt[c-1]c$ $c>=2$
The root can be taken to be integer if c-1 = 1 or c =2 giving b = 2
So solution set $(1,b,c)$ or $(a,2,2)$ or $(a,b,1)$ where a,b,c are any integers
