Question

Quadrilateral ABCD is inscribed in circle 0. Chords BA and CD are extended to intersect at point E. A tangent at B intersects line DA where line
DA is extended to point F. Diagonals BD and AC of quadrilateral ABCD are drawn.

arch ĀB =128
arch BC =144°
arch DC = 64
arch DA = 32

Find the measure of angle 1,2,5 and 6​

Answer 1

Check the picture below.

let's notice the "white" ∡1 is an inscribed angle with an intercepted arc of (x-32), and the "green" ∡5 is also an inscribed angle with an intercepted arc of (2x).

the ∡6 and ∡2 are both external angles, however they intercepted two arcs, a "far arc" and a "near arc", thus we'll use the far arc - near arc formula, as you see in the picture, and we'll use the inscribed angle theorem for the other two.

$\bf \measuredangle 1=\cfrac{x-32}{2}\implies \measuredangle 1 =\cfrac{32}{2}\implies \measuredangle 1 = 16 \\\\[-0.35em] ~\dotfill\\\\ \measuredangle 5 =\cfrac{2x}{2}\implies \measuredangle 5 = x\implies \measuredangle 5 = 64 \\\\[-0.35em] ~\dotfill$

$\bf \measuredangle 2 = \cfrac{(2x+8)~~-~~(x-32)}{2}\implies \measuredangle 2=\cfrac{2x+8-x+32}{2} \\\\\\ \measuredangle 2=\cfrac{x+40}{2}\implies \measuredangle 2=\cfrac{104}{2}\implies \measuredangle 2=52 \\\\[-0.35em] ~\dotfill\\\\ \measuredangle 6=\cfrac{[(2x+8)+(x)]~~-~~(2x)}{2}\implies \measuredangle 6=\cfrac{3x+8-2x}{2}\implies \measuredangle 6=\cfrac{x+8}{2} \\\\\\ \measuredangle 6=\cfrac{72}{2}\implies \measuredangle 6=36$