Given: XZ/MZ=YZ/NZ Prove XZ/MZ=XY/MN

By trigonometric ratio;

XZ/sin(b) = XY/sin(c) = YZ/sin(a)

MN/sin(c) = NZ/sin(f) = MZ/sin(d)

∴ XZ/MZ = XY/MN × (sin(b)/(sin(d))

XY/MN  = XZ/MZ × (sin(b)/sin(d))

Where (sin(b)/(sin(d)) = V;

(XZ/MZ)/(XY/MN) = (XY/MN × V)/(XZ/MZ × V)

(XZ/MZ)/(XY/MN) = (XY/MN)/(XZ/MZ)

∴ (XZ/MZ)² = (XY/MN)²

∴ XZ/MZ = XY/MN QED

Step-by-step explanation:

Whereby the shapes in the question are triangles ΔXYZ and ΔMNZ, and given that we have;

XZ/MZ = YZ/NZ, and from the attached drawing, we have;

∠XZY = ∠NZM

By trigonometric ratio, we have;

XZ/sin(b) = XY/sin(c) = YZ/sin(a)

∴ XZ = sin(b)×XY/sin(c)

MN/sin(c) = NZ/sin(f) = MZ/sin(d)

MZ = sin(d)×MN/sin(c)

∴ XZ/MZ = sin(b)×XY/sin(c)/(sin(d)×MN/sin(c)) = XY/MN × (sin(b)/(sin(d))

XY = sin(c) × XZ/sin(b), MN = sin(c) × MZ/sin(d)

XY/MN  = sin(c) × XZ/sin(b)/(sin(c) × MZ/sin(d)) = XZ/MZ × (sin(b)/sin(d))

Let 'V' represent (sin(b)/(sin(d)), we have;

XZ/MZ = XY/MN × V...(1)

XY/MN = XZ/MZ × V...(2)

Dividing equation (1) by (2) gives;

(XZ/MZ)/(XY/MN) = (XY/MN × V)/(XZ/MZ × V) = (XY/MN)/(XZ/MZ)

(XZ/MZ)/(XY/MN) = (XY/MN)/(XZ/MZ)

∴ (XZ/MZ)² = (XY/MN)²

∴ XZ/MZ = XY/MN

last one.

step-by-step explanation:

last one.

generally speaking, the others follow some sort of pattern that correlates to either linear (looks like a straight line) or sinusoidal (looks like an s).

the last one is just a random cloud of points that doesn't correlate to any particular known order.

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58