Proof:

Step 1: Join OA, OB, OC

Since tangents from external point are equal:

AP = AC (from point A)

BQ = BC (from point B)

Step 2: In ΔAPO and ΔACO:

AP = AC (tangents from A)

OP = OC (radii)

OA = OA (common)

∴ ΔAPO ≅ ΔACO (by SSS)

⇒ ∠POA = ∠COA …(1)

Step 3: Similarly, ΔBQO ≅ ΔBCO

⇒ ∠QOB = ∠COB …(2)

Step 4: Since XY ∥ X’Y’, points P, O, Q are collinear

∠POQ = 180° (straight line)

Step 5: ∠POQ = ∠POA + ∠AOC + ∠COB + ∠BOQ

180° = ∠COA + ∠COA + ∠COB + ∠COB (from 1 & 2)

180° = 2(∠COA + ∠COB)

180° = 2∠AOB

∠AOB = 90°

Proof:

Step 1: PA = PB (tangents from external point)

In ΔPAB, PA = PB ⇒ ∠PAB = ∠PBA = x (angles opposite equal sides)

Step 2: In ΔPAB:

∠PAB + ∠PBA + ∠APB = 180°

x + x + ∠APB = 180°

∠APB = 180° – 2x …(1)

Step 3: Radius is perpendicular to tangent at point of contact

OA ⊥ PA ⇒ ∠OAP = 90°

∠PAB + ∠OAB = 90°

x + ∠OAB = 90°

x = 90° – ∠OAB …(2)

Step 4: Substitute (2) in (1):

∠APB = 180° – 2(90° – ∠OAB)

∠APB = 180° – 180° + 2∠OAB

∠APB = 2∠OAB

Solution:

Step 1: Find HCF of 36, 60, and 84

36 = 2² × 3²

60 = 2² × 3 × 5

84 = 2² × 3 × 7

HCF = 2² × 3 = 4 × 3 = 12

Step 2: Maximum teachers per room = HCF = 12

Step 3: Number of rooms required:

For English teachers: 36 ÷ 12 = 3 rooms

For Hindi teachers: 60 ÷ 12 = 5 rooms

For Science teachers: 84 ÷ 12 = 7 rooms

Step 4: Total rooms = 3 + 5 + 7 = 15

Solution:

Step 1: Find zeroes by factorization

\(2x^2 – (1 + 2\sqrt{2})x + \sqrt{2}\)

= \(2x^2 – x – 2\sqrt{2}x + \sqrt{2}\)

= \(x(2x – 1) – \sqrt{2}(2x – 1)\)

= \((2x – 1)(x – \sqrt{2})\)

Step 2: Zeroes are:

\(2x – 1 = 0\) ⇒ \(x = \frac{1}{2}\)

\(x – \sqrt{2} = 0\) ⇒ \(x = \sqrt{2}\)

Step 3: Verification with coefficients:

For \(ax^2 + bx + c = 0\):

Sum of zeroes = \(-b/a\)

Product of zeroes = \(c/a\)

Step 4: Calculate sum:

\(\alpha + \beta = \frac{1}{2} + \sqrt{2} = \frac{1 + 2\sqrt{2}}{2}\)

\(-b/a = -\frac{[-(1 + 2\sqrt{2})]}{2} = \frac{1 + 2\sqrt{2}}{2}\) ✓

Step 5: Calculate product:

\(\alpha \times \beta = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2}\)

\(c/a = \frac{\sqrt{2}}{2}\) ✓

Proof:

Given: \(\sin\theta + \cos\theta = \sqrt{3}\)

Step 1: Square both sides:

\((\sin\theta + \cos\theta)^2 = (\sqrt{3})^2\)

\(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 3\)

Step 2: Use identity \(\sin^2\theta + \cos^2\theta = 1\):

\(1 + 2\sin\theta\cos\theta = 3\)

\(2\sin\theta\cos\theta = 2\)

\(\sin\theta\cos\theta = 1\) …(1)

Step 3: Find \(\tan\theta + \cot\theta\):

\(\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\)

= \(\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}\)

Step 4: Substitute from (1):

= \(\frac{1}{1} = 1\)

Proof:

Prove that:

\[\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A\]

Proof:

Step 1: Divide numerator and denominator of LHS by \(\sin A\):

\[
LHS = \frac{\frac{\cos A}{\sin A} – \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} – \frac{1}{\sin A}}
\]

\[
= \frac{\cot A – 1 + \csc A}{\cot A + 1 – \csc A}
\]

Step 2: Rearrange terms:

\[
LHS = \frac{(\cot A + \csc A) – 1}{(\cot A – \csc A) + 1}
\]

Step 3: Use identity \(1 = \csc^2 A – \cot^2 A\):

\[
LHS = \frac{(\cot A + \csc A) – (\csc^2 A – \cot^2 A)}{1 + (\cot A – \csc A)}
\]

Step 4: Factor difference of squares:

\[
\csc^2 A – \cot^2 A = (\csc A – \cot A)(\csc A + \cot A)
\]

\[
= \frac{(\cot A + \csc A) – (\csc A – \cot A)(\csc A + \cot A)}{1 – (\csc A – \cot A)}
\]

Step 5: Factor common term:

\[
= \frac{(\cot A + \csc A)[1 – (\csc A – \cot A)]}{1 – (\csc A – \cot A)}
\]

Step 6: Cancel common factor:

\[
LHS = \cot A + \csc A
\]

\[
\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A
\]

Hence proved.

Solution:

Problem:

A coin is flipped 3 times. Vidhi drives if she gets two heads in a row (HH pattern). Unnati drives if she gets H immediately followed by T (HT pattern). Who has a greater probability to drive?

Total Possible Outcomes:

Number of coin flips: 3

Total outcomes: \( 2^3 = 8 \)

Sample space: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Case 1: Vidhi drives (two heads in a row – HH pattern)

Favorable outcomes:

• HHH
• HHT
• THH

Number of favorable outcomes: 3

\[P(\text{Vidhi}) = \frac{3}{8} = 0.375\]

Case 2: Unnati drives (H followed by T – HT pattern)

Favorable outcomes:

• HHT
• HTH
• HTT
• THT

Number of favorable outcomes: 4

\[P(\text{Unnati}) = \frac{4}{8} = 0.5\]

Comparison:

\[P(\text{Vidhi}) = \frac{3}{8} = 0.375\]

\[P(\text{Unnati}) = \frac{4}{8} = 0.5\]

Since \( 0.5 > 0.375 \),

\[P(\text{Unnati}) > P(\text{Vidhi})\]

Conclusion:

Unnati has a greater probability to drive the car.

Solution for (A):

Step 1: Let incomes be 3x and 4x

Let expenditures be 5y and 7y

Step 2: Savings = Income – Expenditure

For Aryan: 3x – 5y = 15000 …(1)

For Babban: 4x – 7y = 15000 …(2)

Step 3: Solve equations:

Multiply (1) by 4: 12x – 20y = 60000 …(3)

Multiply (2) by 3: 12x – 21y = 45000 …(4)

Step 4: Subtract (4) from (3):

Step 4: Subtract Equation (4) from (3)

(12x – 20y) – (12x – 21y) = 60000 – 45000

12x – 20y – 12x + 21y = 15000

y = 15000

Step 5: Find x

Substitute y = 15000 in equation (1):

3x – 5(15000) = 15000

3x – 75000 = 15000

3x = 90000

x = 30000

Step 6: Find Incomes

Aryan’s income = 3x = 3 × 30000 = ₹90000

Babban’s income = 4x = 4 × 30000 = ₹120000

Step 7: Verification

Aryan’s expenditure = 5y = 5 × 15000 = ₹75000

Aryan’s savings = 90000 – 75000 = ₹15000 ✓

Babban’s expenditure = 7y = 7 × 15000 = ₹105000

Babban’s savings = 120000 – 105000 = ₹15000 ✓

Answer:

Aryan’s income = ₹90000

Babban’s income = ₹120000

Solution

1. Solve the system of equations graphically and find the area of the triangle formed by the lines and the x-axis.

Step 1: Plot 2x + y = 6
Points: (0,6) and (3,0)

Step 2: Plot 2x − y − 2 = 0
Points: (0,−2) and (1,0)

Step 3: Find intersection
Adding the equations: 4x − 2 = 6 → 4x = 8 → x = 2
Substitute into first: 2(2) + y = 6 → y = 2
Intersection point: (2,2)

Step 4: Triangle vertices
A(1,0), B(3,0), C(2,2)

Step 5: Area
Base = 3 − 1 = 2
Height = 2
Area = ½ × 2 × 2 = 2

Answer:

Solution: x = 2, y = 2
Area = 2 square units

Proof:

Age Problem Solution

Let the present age of the Father age be x years and that of his son be y years.

Step 1: Equation from conditions after 5 years

After 5 years Father’s age = x + 5

After 5 years son’s age = y + 5

As per the question:

x + 5 = 3(y + 5)

⇒ x – 3y = 10 …(i)

Step 2: Equation from conditions 5 years ago

5 years ago Father’s age = x – 5

5 years ago son’s age = y – 5

As per the question:

x – 5 = 7(y – 5)

⇒ x – 7y = -30 …(ii)

Step 3: Solve the equations

Subtracting (ii) from (i):

(x – 3y) – (x – 7y) = 10 – (-30)

⇒ 4y = 40

⇒ y = 10

Step 4: Find x

Putting y = 10 in equation (i):

x – 3 × 10 = 10

⇒ x = 10 + 30 = 40

Answer:

Man’s present age = 40 years

Son’s present age = 10 years