According to question,
Given :
BAC is right angle triangle
AB = $$\frac{1}{2}$$BC

$$\eqalign{ & \frac{{AB}}{{BC}} = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{P}{H} = \frac{1}{2} \cr & \sin \theta = \frac{P}{H} = \frac{1}{2} \cr & \sin \,{30^ \circ } = \frac{1}{2} \cr & \therefore \theta = \angle ACB = {30^ \circ } \cr} $$

According to question,

Area of Δ(OBA + OAC + OBC) = area of ΔABC
$$\frac{1}{2}$$ × 4x × 3 + $$\frac{1}{2}$$ × 5x × 3 + $$\frac{1}{2}$$ × 6x × 3 = $$\frac{1}{2}$$ × (6x × AD)
$$\frac{1}{2}$$ × 3(4x + 5x + 6x) = $$\frac{1}{2}$$ × (6x × AD)
45x = 6x × AD
AD = $$\frac{{15}}{2}$$
AD = 7.5 cm

According to question,

In ΔPQR
x + y + 40° = 180°
x + y = 140° . . . . . . . (i)
In ΔAQC
x + x + ∠C = 180°
∠C = 180° - 2x . . . . . . . . (ii)
In ΔBCR
y + y + ∠C = 180°
∠C = 180° - 2y . . . . . . . . (iii)
But ∠ACB = 180° - 180° + 2x - 180° +2y
∠ACB = 2x + 2y - 180°
∠ACB = 2(x + y) - 180° . . . . . . . . . (iv)
Put the value of equation (i) in equation (iv)
∠ACB = 2 × 140° - 180°
∠ACB = 280° - 180°
∠ACB = 100°

According to question,
Given :
∠ACB = 55°
∠ABC = 65°
∠BIC = ?

∴ ∠ACB + ∠ABC + ∠BAC = 180°
∠BAC = 180° - 55° - 65°
∠BAC = 60°
We know that
∠BIC = 90 + $$\frac{1}{2}$$ ∠A
∠BIC = 90 + $$\frac{1}{2}$$ × 60
∠BIC = 90 + 30
∠BIC = 120°

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Alternate:
In ΔBIC,
$$\frac{1}{2}$$ ∠B + $$\frac{1}{2}$$ ∠C + ∠BIC = 180°
$$\frac{1}{2}$$ (65° + 55°) + ∠BIC = 180°
∠BIC = 180° - 60°
∠BIC = 120°

According to question,
ABC is a right angle triangle

∴ Apply Pythagoras theorem
AC² = AB² +BC²
(x + 2)² = (x - 2)² + x²
x² + 4 + 4x = x² + 4 - 4x + x²
x² = 8x
x = 8

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Alternate : From option approach
$$\eqalign{ & \left( {x - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x + 2} \right) \cr & \,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Triplet}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x}} = {\text{8}} \cr} $$

According to question,
Given : ∠ABC = 90°
$$\frac{{AN}}{{NC}} = ?$$

ΔABC ∼ ΔBNC
ΔABC ∼ ΔANB
∴ ΔABC ∼ ΔBNC ∼ ΔANB
AB = 3, BC = 4, AC = 5

$$\eqalign{ & \frac{{AB}}{{BN}} = \frac{{AC}}{{BC}} \cr & BN = \frac{{AB \times BC}}{{AC}} = \frac{{3 \times 4}}{5} = 2.4 \cr & \frac{{BC}}{{NC}} = \frac{{AB}}{{NB}} \cr & \frac{4}{{NC}} = \frac{3}{{2.4}} \cr & NC = 3.2 \cr & \frac{{AB}}{{AN}} = \frac{{BC}}{{NB}}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{{AN}} = \frac{4}{{2.4}} \cr & AN = 1.8 \cr & \frac{{AN}}{{NC}} = \frac{{1.8}}{{3.2}} \cr & \frac{{AN}}{{NC}} = \frac{9}{{16}} \cr & \therefore AN:NC = 9:16 \cr} $$

According to question,
ABC is a right angle triangle,

∴ AB = 8 cm
   BC = 6 cm
∴ AC² = AB² + BC²
   AC = 64 + 36
   AC = $$\sqrt {100} $$
   AC = 10 cm
In right triangle
Circum Radius I_R = $$\frac{{AC}}{2}$$ = $$\frac{{10}}{2}$$ = 5 cm

According to question,
Given : ∠BOC = 110°
∠BOC = 90° + $$\frac{1}{2}$$ ∠A

110° = 90° + $$\frac{{\angle A}}{2}$$
$$\frac{{\angle A}}{2}$$ = 20
∠A = 40°

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

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

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

According to question,

In equilateral triangle 'AD' bisects the BC in two equal parts.
Let side of equilateral triangle is 2 cm
$$\eqalign{ & \therefore \frac{{AB}}{{BD}} = \frac{2}{1} \cr & AB:BD = 2:1 \cr} $$

According to question,

Let AB = AC = 2x
∵ AQ = QC = x
∴ AB is a secant
∴ AP × AB = AQ²
AP × 2x = x²
$$\eqalign{ & AP = \frac{x}{2} \cr & \frac{{AP}}{{AB}} = \frac{x}{{2 \times 2x}} = \frac{1}{4} \cr & \frac{{AP}}{{AB}} = \frac{1}{4} \cr & AP:AB = 1:4 \cr} $$

According to question,

Given :
∠D = 50°
∠BAC = 2∠BDC (property)
∴ ∠BAC = 2 × 50°
∠BAC = 100°

According to question,

AD is the median and 'C' is the centroid
∴ AO = 10 cm
2 units = 10
1 units = 5
∴ OD = 5 cm

According to question,

AB + BC = 12 cm
BC + CA = 14 cm
CA + AB = 18 cm
2(AB + BC + CA) = 44 cm
AB + BC + CA = $$\frac{{44}}{2}$$ cm
AB + BC + CA = 22 cm
Perimeter of triangle = 22 cm
Perimeter of triangle = Perimeter of circle
22 = 2πr
2 × $$\frac{{22}}{7}$$ × r = 22
r = $$\frac{7}{2}$$ cm

ar ADE = ar DEBC
So, ar ΔADE = 1 unit² and ar ABC = 2 unit²
$$\eqalign{ & \frac{{{\text{ar}}\,\Delta ADE}}{{{\text{ar}}\,\Delta ABC}} = \frac{{A{D^2}}}{{A{B^2}}} \cr & \frac{1}{2} = {\left( {\frac{{AD}}{{AB}}} \right)^2} \cr & \frac{1}{{\sqrt 2 }} = \frac{{AD}}{{AB}} \cr & \therefore \frac{{AD}}{{DB}} = \frac{1}{{\sqrt 2 - 1}} \cr & \left( {\therefore DB = AB - AD = \sqrt 2 - 1} \right) \cr & {\text{So,}}\,AD:BD = 1:\sqrt 2 - 1 \cr} $$

According to question,
Given:

∴ ∠BOC = 90° + $$\frac{1}{2}$$ ∠A
∠BOC = 90° + $$\frac{1}{2}$$ × 30°
∠BOC = 90° + 15°
∠BOC = 105°

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°

According to question,

Given :
∠A = 40°       AB = AC
∴ ∠B = ∠C
In ΔABC
∠A + ∠B + ∠C = 180°
40° + 2∠B = 180°
2∠B = 180° - 40°
2∠B = 140°
∠B = 70°
∴ External angle at B = 180° - 70° = 110°

According to question,
Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle.

Note: In right angle triangle median divides the hypotenuse in two equal parts
∴ BD = $$\frac{{\text{H}}}{2}$$
BD = $$\frac{{10}}{2}$$
BD = 5 cm

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

According to question,
If the median drawn on the base of a triangle is half of its base of the triangle then the triangle will be right angled triangle.

According to question,

Given: BAC is a right angle triangle
AD ⊥ BC
AD = 6 cm
BD = 4 cm
BC = ?
In ΔBAD
$$\eqalign{ & AB = \sqrt {B{D^2} + A{D^2}} \cr & AB = \sqrt {{4^2} + {6^2}} \cr & AB = \sqrt {52} \,cm \cr} $$
ΔBAC ∼ ΔBDA


$$\eqalign{ & \therefore \frac{{BC}}{{AB}} = \frac{{AB}}{{BD}} \cr & \therefore \frac{{BC}}{{\sqrt {52} }} = \frac{{\sqrt {52} }}{4} \cr & BC = \frac{{52}}{4} \cr & BC = 13\,cm \cr} $$

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Alternate :
$$\eqalign{ & A{B^2} = BD.BC \cr & {\left( {\sqrt {B{D^2} + A{D^2}} } \right)^2} = BD.BC \cr & {\left( {\sqrt {{4^2} + {6^2}} } \right)^2} = 4.BC \cr & \frac{{52}}{4} = BC, \cr & \therefore BC = 13\,cm \cr} $$

According to question,

Given :
∠ABC = 60°
∠BCA = 80°
∠BIC = ?
∠BAC = 40°
∴ ∠BIC = 90° + $$\frac{1}{2}$$ × 40°
∠BIC = 110°

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°

According to question,
Given:
∠BAC = 85°
∠BCA = 75°
∠OAC = ?

∠ABC + ∠BCA + ∠CAB = 180°
∠ABC = 20°
∴ ∠COA = 2 × ∠ABC
    ∠COA = 2 × 20 = 40°
In ΔAOC
We know OC = OA
∴ ∠OAC = ∠OCA
∴ ∠OAC + ∠OCA + ∠COA = 180°
2∠OAC = 180° - 40°
2∠OAC = 140°
∠OAC = 70°

According to question,

ΔABC ∼ ΔACD
$$\eqalign{ & \therefore \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{B{C^2}}}{{A{D^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{{8^2}}}{{{6^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{64}}{{36}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{16}}{9} \cr} $$
area ΔABC : area ΔACD = 16 : 9

According to question,

Given : ∠BOD = 15°
∴ ∠BDO + ∠DOB + ∠DBO = 180°
∠DBO = 75°
∠ABC = 2 × ∠DBO
∠ABC = 2 × 75°
∠ABC = 150°

According to question,
Let sides of the triangle be 3x, 4x, 6x
Now check the square of biggest side and sum of square of two smallest side and check which is greater.
∴ (3x)² + (4x)² < (6x)²
⇒ 25x² < 36x²
∴ The triangle will be obtuse angled triangle.