Question

solve question no 17 it is a good question

Answer 1

Question:-

$\sf{if\:angles\:A, B, c\:of\:a\: \triangle ABC\:forms}\\ \sf{an\:increasing \:AP\: ,Then, \sin B=}}$

$\rule{200}{2}$

Answer :-

If the AP is an increasing AP means that the common difference will always be positive.

Now we know that :-

If x, y and z are in AP, then,

$2y = x +z$

$\rule{200}{2}$

Similarly,

In triangle ABC :-

Angles A, B and C are in AP

Therefore :-

$\boxed{2 \mathbb{B=A+C}}$

$\rule{200}{2}$

Now,

$\sin(b) = \sin( \frac{a + c}{2} )$

....... (i)

$\rule{200}{2}$

The angle sum. Property of a triangle is equal to 180 degrees.

$a + b + c = 180 \\ \\ b = 180 - (a + c)$

Also,

$b = \frac{a + c}{2}$

Thus,

$\frac{a + c}{2} = 180 - (a + c) \\ \\ \\ \frac{a + c}{2} + (a + c) = 180 \\ \\ \\ 3a + 3c = 360 \\ \\ \\ 3(a + c) = 360 \\ \\ \\ a + c = \frac{360}{3} = 120$

$\rule{200}{2}$

Now, back to ..... (i)

We have

$\begin{aligned} \sin(b) =\sin(\frac{a+c}{2}) \\ \\ =\sin(\frac{120}{2}) \\ \\ =\sin(60) \end{aligned}$

$\rule{200}{2}$

We know that

$\sin(60) = \frac{ \sqrt{3}}{2}$

Therefore :-

$\sin(b) = \sin(60) = \frac{ \sqrt{3}}{2}$

Option (b)