Solution:
According to question,
Given: ∠C = 90°

BC = AC = 5 cm (Isosceles triangle)
By Pythagoras theorem
AB² = AC² + BC²
AB² = 5² + 5²
AB² = 25 + 25
AB² = 50
AB = 5$$\sqrt 2 $$ cm

Solution:
According to question,
Given : ABC is a right angle triangle

BC = 6$$\sqrt 3 $$
∴ sin 30° = $$\frac{P}{H}$$ = $$\frac{{AB}}{{6\sqrt 3 }}$$
AB = 3$${\sqrt 3 }$$

sin 60° = $$\frac{P}{H}$$ = $$\frac{{AN}}{{AB}}$$
$$\frac{{\sqrt 3 }}{2}$$ = $$\frac{{AN}}{{3\sqrt 3 }}$$
AN = $$\frac{9}{2}$$
AN = 4.5 cm

Solution:
According to question,

PX || BD [mid point theorem]
∴ PX = $$\frac{1}{2}$$BD
Similarly, QY || BD
∴ QY = $$\frac{1}{2}$$BD
∴ PX : QY,           $$\frac{1}{2}$$BD : $$\frac{1}{2}$$BD
PX : QY = 1 : 1

Solution:
According to question,
Given :

AB = AC = 2a
BC = a
AD ⊥ BC
In isosceles triangle perpendicular sides bisects the opposite side of the length
∴ BD = $$\frac{{BC}}{2}$$
   BD = $$\frac{a}{2}$$
In ΔADB using Pythagoras theorem
AB² = BD² + AD²
(2a)² = $${\left( {\frac{a}{2}} \right)^2}$$ + AD²
4a² = $$\frac{{{a^2}}}{4}$$ + AD²
AD² = 4a² - $$\frac{{{a^2}}}{4}$$
AD = $$\frac{{\sqrt {15} }}{2}$$ a units

Solution:

AD² = BD × DC
AD² = 5 × 9
AD = $$\sqrt {45} $$
AD = $$3\sqrt 5 $$  cm

Solution:
According to question,

Given : ∠BOC = 60°
As we know that
∴ ∠O = 90 - $$\frac{1}{2}$$ ∠A
⇒ $$\frac{1}{2}$$ ∠A = 90° - 60°
⇒ $$\frac{1}{2}$$ ∠A = 30°
⇒ ∠A = 60°

Solution:
According to question,
Given :

PQR is an equilateral triangle
QR = RS
PR = RS
∠SRP = 180° - 60° (Exterior ∠)
∠SRP = 120°
∴ ∠RPS = ∠RSP
∴ ∠RPS + ∠PRS + ∠RSP = 180°
2∠PSR = 180° - 120°
∠PSR = $$\frac{{{{60}^ \circ }}}{2}$$
∠PSR = 30°