Answer 1

We know that `cos theta = "𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒"/"ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒"`

Let us consider right angled Δ^le ABC

Let x be the opposite side, By applying Pythagoras theorem

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

25 = 𝑥² + 9

𝑥² = 16 ⇒ 𝑥 = 4

`sin theta = (AB)/(AC) = 4/5`

`tan theta = (AB)/(BC) = 4/3`

Substitute sin 𝜃, tan 𝜃 in equation we get

`(sin theta 1/(tan theta))/(2 tan theta) = (4/5 3/4)/(2 xx 4/3)`

`= ((16 - 15)/20)/(8/3) = (1/20)/(8/3)`

`=  1/20 xx  3/4 = 3/160`