Here, ΔABC ～ ΔADE (given)
∴ Their corresponding sides are proportional (by RHS similarity).
∴ \(\frac { AC }{ AE } =\frac { BC }{ DE } \)
⇒ \(\frac { 12 }{ 96 } =\frac { 54 }{ h } \)
⇒ h = \(\frac { 5\times 96 }{ 12 } \) = 40 feet
∴ The height of the tree is 40 feet.

Given that ΔPEN ～ ΔPAD,
∴ \(\frac { PE }{ PA } =\frac { EN }{ AD } \Rightarrow \frac { 4 }{ 7 } =\frac { 6 }{ x } \Rightarrow x=\frac { 42 }{ 4 } \)
Also, \(\frac { PE }{ PA } =\frac { PN }{ PD } =\frac { y }{ y+5 } \)
i.e. \(\frac { 4 }{ 7 } =\frac { y }{ y+5 } \) ⇒ 4y + 20 =7y ⇒ 7y - 4y = 20
3y = 20 ⇒ \(y=\frac { 20 }{ 3 } \)cm

Given that  ΔPQR ～ ΔXYZ
∴ Their corresponding sides are proportional
⇒ \(\frac { PQ }{ XY } =\frac { QR }{ YZ } =\frac { PR }{ XZ } \)
⇒ \(\frac { 8 }{ a } =\frac { 14 }{ b } =\frac { 10 }{ 16 } \)
⇒ \(\frac { 8 }{ a } =\frac { 10 }{ 16 } \)
⇒ \(a=\frac { 8\times 16 }{ 10 } =\frac { 128 }{ 10 } \)
a = 12.8 cm
Also, \(\frac { 14 }{ b } =\frac { 10 }{ 16 } \)
⇒ \(\frac { 14\times 16 }{ 10 } =\frac { 224 }{ 10 } \)
∴ b = 22.4 cm

Now, from Fig \(\angle\)140^o+ \(\angle\)z = \(\angle\)180^o (linear pair)
⇒ \(\angle\)z = 180^o −140^o = 40^o
Also \(\angle\)x+ \(\angle\)z= \(\angle\)70 + \(\angle\)z (exterior angle property)
⇒\(\angle\)x=70^o
Also \(\angle\)z+ \(\angle\)y + 70^o = 180^o (angle sum property in ΔABC)
⇒ 40^o + \(\angle\)y + 70^o = 180^o
⇒ \(\angle\)y = 180^o − 110^o = 70^o

Extend MO and let it meet RS at Y. Extend NO and let it meet PQ at X.
As PQ || RS,
∠OYN = ∠OMX = 55^o (alternate angles are equal)
∴ ∠MON = ∠OYN +∠ONY (exterior angle of ΔOYN = sum of interior opposite angles)
= 55^o + 30^o = 85^o
⇒∠MOX = 180^o− 85^o = 95^o (∠MON, ∠MOX are linear pair)