Question

Plz answer it I want to get a b at least

Answer 1

Δ$WPX$,  $PX = 6, WX = 12$

$b^2 + s^2 = h^2\\\\b^2 = 12^2 - 6^2\\\\b^2 = 144 - 36\\\\b^2 = 108\\\\b = \sqrt{108}\\\\b = 6\sqrt{3}$

Now, sinX = $\frac{b}{XW}$

= $\frac{6\sqrt{3}}{12}$

= $\frac{\sqrt{3}}{2}$
And as we know, sin60* = $\frac{\sqrt{3}}{2}$

So, ∠X = 60*
Therefore, ∠W = 30* (Angle sum property)

Now, sin(W) = $\frac{ZQ}{WZ}$ = sin30 = $\frac{1}{2}$

$\frac{ZQ}{WZ} = \frac{1}{2}\\\\\frac{ZQ}{18} = \frac{1}{2}\\\\ZQ = \frac{18}{2}\\\\ZQ = 9$
So, ZQ = 9

D) In △AOM and △BOM,

∠AMO = ∠BMO (right angles)
AO = BO  (radius)
MO = MO (common)

So, by RHS criterion, △AOM ≅ △BOM

Therefore, AM = BM (Common Parts of Congruent Triangles)