Step 1: Recall the Definition of Tangent
\[ \tan \theta = \frac{Opposite}{Adjacent} \]
Since \(\tan \theta = \tfrac{3}{4}\), this means:
- Opposite (Perpendicular) = 3 units

- Adjacent (Base) = 4 units


Step 2: Apply the Pythagorean Theorem to Find the Hypotenuse
\[ Hypotenuse^2 = Opposite^2 + Adjacent^2 \] \[ Hypotenuse^2 = 3^2 + 4^2 = 9 + 16 = 25 \] \[ Hypotenuse = \sqrt{25} = 5 \]

Step 3: Formula for Cosine
\[ \cos \theta = \frac{Adjacent}{Hypotenuse} = \frac{4}{5} \]

Step 4: Conclusion

The value of \(\cos \theta\) is \(\tfrac{4}{5}\), which corresponds to option (A).

\[ \boxed{\cos \theta = \tfrac{4}{5}} \] Quick Tip: Whenever \(\tan \theta\) is provided, visualize a right triangle where the opposite side corresponds to the numerator and the adjacent side to the denominator. Then, use the Pythagorean theorem to find the hypotenuse and proceed to compute \(\sin \theta\) or \(\cos \theta\).

Step 1: Total Possible Outcomes

When a fair die is rolled once, the possible outcomes are \{1, 2, 3, 4, 5, 6\.

Thus, the total number of outcomes is 6.


Step 2: Favorable Outcomes for an Even Number

The even numbers on the die are \{2, 4, 6\, so the number of favorable outcomes is 3.


Step 3: Applying the Probability Formula

The probability of an event is given by:
\[ P(Event) = \frac{Number of favorable outcomes}{Total number of outcomes} \]
For this case:
\[ P(even number) = \frac{3}{6} = \frac{1}{2} \]

Step 4: Conclusion

The probability of rolling an even number is \(\frac{1}{2}\).

Thus, the correct answer is option (B).
Quick Tip: For probability questions, first identify the total possible outcomes, then determine the favorable ones. Applying this formula gives the desired probability.

Step 1: Determine the Total Frequency and Median Position

The total frequency is: \( N = 7 + 8 + 15 + 10 + 5 = 45 \).

The median position is calculated as: \( \frac{N}{2} = \frac{45}{2} = 22.5 \).


Step 2: Calculate Cumulative Frequencies (cf)

- For \(0\)--\(10\): \( 7 \)

- For \(10\)--\(20\): \( 7 + 8 = 15 \)

- For \(20\)--\(30\): \( 15 + 15 = 30 \)

- For \(30\)--\(40\): \( 30 + 10 = 40 \)

- For \(40\)--\(50\): \( 40 + 5 = 45 \)


Step 3: Identify the Median Class

We need to find the first class interval where the cumulative frequency is greater than or equal to \(22.5\).

From the cumulative frequency table, the first class whose cumulative frequency exceeds 22.5 is \(20\(--\)30\) with a cumulative frequency of 30.

Thus, the median class is \(20\(--\)30\).

\[ \boxed{Median class =\, 20--30} \] Quick Tip: To determine the median class, calculate \( \frac{N}{2} \) and find the class whose cumulative frequency is just greater than or equal to \( \frac{N}{2} \). No complex formula is needed for this step.

Step 1: Identify the Highest Frequency

The frequencies for the class intervals are \(2,\, 7,\, 11,\, 8,\, 6\). The highest frequency is \(11\).


Step 2: Determine the Modal Class

The class interval with the maximum frequency \(11\) is \(10\(--\)15\), so this is the modal class.

\[ \boxed{Modal class =\, 10--15 \] Quick Tip: To find the \emph{modal class} in grouped data, simply identify the class with the highest frequency. This method works for most cases where you are not asked to calculate the exact mode.

Step 1: Understanding the Range of Probability

The probability of an event must always be between 0 and 1, inclusive. Therefore, we have the condition: \[ 0 \leq P(E) \leq 1 \]

Step 2: Analyzing the Options

- Option (A) \(1\): Valid probability, indicating a certain event.

- Option (B) \(-1\): Not possible, as probability cannot be negative.

- Option (C) \(\frac{1}{2}\): A valid probability, representing a 50% chance.

- Option (D) \(\frac{1}{3}\): A valid probability, representing a one-third chance.


Step 3: Conclusion

Since probabilities cannot be negative, \(-1\) is not a valid probability value.


Thus, the correct answer is option (B).
Quick Tip: Always remember: The range for any probability is between 0 and 1, inclusive. Any value less than 0 or greater than 1 is invalid.

Step 1: Express \(\cot A\)

We are given \(3 \cot A = 4 \Rightarrow \cot A = \frac{4}{3}\).

Thus, \(\tan A = \frac{1}{\cot A} = \frac{3}{4}\).


Step 2: Substitute into the Expression

We need to evaluate: \[ \dfrac{1 - \tan^2 A}{1 + \tan^2 A} \]

Substitute \(\tan A = \frac{3}{4}\): \[ \tan^2 A = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

Now, substitute into the expression: \[ \dfrac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \dfrac{\frac{16}{16} - \frac{9}{16}}{\frac{16}{16} + \frac{9}{16}} = \dfrac{\frac{7}{16}}{\frac{25}{16}} = \dfrac{7}{25} \]

Step 3: Applying Trigonometric Identity

We recognize that: \[ \dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos 2A \]

Since \(\tan A = \frac{3}{4}\), the angle \(A\) is acute, meaning \(0 < A < \frac{\pi}{2}\).
Thus, \(2A\) lies in the second quadrant, where \(\cos 2A\) is negative. Therefore, the value of \(\cos 2A\) is \(-\frac{7}{25}\).


Step 4: Conclusion

The value of \(\dfrac{1 - \tan^2 A}{1 + \tan^2 A}\) is \(-\frac{7}{25}\).

Thus, the correct answer is option (B).
Quick Tip: When dealing with trigonometric identities, always consider the quadrant of the angle to determine the correct sign for values like \(\cos 2A\).

Step 1: Recall standard trigonometric values

From the trigonometric table:
\[ \cos 0^\circ = 1,\ \cos 30^\circ = \frac{\sqrt{3}}{2},\ \cos 45^\circ = \frac{1}{\sqrt{2}},\ \cos 60^\circ = \frac{1}{2},\ \cos 90^\circ = 0 \]

Step 2: Apply for \(\cos 60^\circ\)

Clearly, \[ \cos 60^\circ = \frac{1}{2} \]
\[ \boxed{\cos 60^\circ = \tfrac{1}{2}} \] Quick Tip: Always remember the standard trigonometric ratios for \(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\). They are frequently asked in exams.

Step 1: Recall Pythagorean identities
\[ 1 + \cot^2 \theta = \csc^2 \theta \quad and \quad 1 + \tan^2 \theta = \sec^2 \theta \]

Step 2: Substitute these identities into the given expression
\[ \frac{1+\cot^2 \theta}{1+\tan^2 \theta} = \frac{\csc^2 \theta}{\sec^2 \theta} \]

Step 3: Simplify the ratio
\[ \frac{\csc^2 \theta}{\sec^2 \theta} = \frac{\dfrac{1}{\sin^2 \theta}}{\dfrac{1}{\cos^2 \theta}} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \]

Wait, let's carefully check:

- Numerator: \(1+\cot^2 \theta = \csc^2 \theta\)
- Denominator: \(1+\tan^2 \theta = \sec^2 \theta\)

So: \[ \frac{\csc^2 \theta}{\sec^2 \theta} = \frac{\dfrac{1}{\sin^2 \theta}}{\dfrac{1}{\cos^2 \theta}} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \]

Step 4: Final Answer
\[ \boxed{\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta} = \cot^2 \theta} \]

Thus, the correct option is (D).
Quick Tip: Whenever you see expressions like \(1+\cot^2 \theta\) or \(1+\tan^2 \theta\), immediately recall the Pythagorean identities: \(1+\cot^2 \theta = \csc^2 \theta\) and \(1+\tan^2 \theta = \sec^2 \theta\).

Step 1: Relation between LCM and HCF

For two numbers \(a\) and \(b\): \[ LCM(a, b) \times HCF(a, b) = a \times b \]

Step 2: Apply the formula

Here \(a = 35\), \(b = 63\), and \(LCM(35, 63) = 315\).

\[ HCF(35, 63) = \dfrac{a \times b}{LCM(a, b)} \]
\[ HCF = \dfrac{35 \times 63}{315} \]
\[ HCF = \dfrac{2205}{315} = 7 \]

Step 3: Conclusion

Thus, the \(HCF(35, 63) = 7\).


The correct answer is option (B).
Quick Tip: Always remember the relation: \(HCF \times LCM = Product\) of the numbers. This is very useful for solving LCM/HCF problems quickly.

Step 1: Prime factorization

- \(12 = 2^2 \times 3\)

- \(15 = 3 \times 5\)

- \(21 = 3 \times 7\)


Step 2: Take highest powers of all primes
\[ LCM = 2^2 \times 3 \times 5 \times 7 \]
\[ LCM = 4 \times 3 \times 5 \times 7 = 420 \]

Step 3: Re-check options

The \(LCM(12, 15, 21) = 420\).


So, the correct answer is option (D).
Quick Tip: For finding \(LCM\), always use the highest powers of all prime factors present in the numbers.

Step 1: Multiply the given terms
\[ \sqrt{2} \times (2-\sqrt{2}) = \sqrt{2} \times 2 - \sqrt{2} \times \sqrt{2} \] \[ = 2\sqrt{2} - 2 \]

Step 2: Classify the result

The expression \(2\sqrt{2} - 2\) is the difference of an irrational number (\(2\sqrt{2}\)) and a rational number (\(2\)).

Thus, the result is irrational.

\[ \boxed{2\sqrt{2} - 2 \ is irrational} \] Quick Tip: Multiplying a surd (like \(\sqrt{2}\)) with a linear expression often produces an irrational result unless special cancellation occurs.

Step 1: Recall the distance formula
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute values

Points are \((x_1,y_1)=(2,3)\) and \((x_2,y_2)=(4,1)\).
\[ d = \sqrt{(4-2)^2 + (1-3)^2} \] \[ = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \]

Step 3: Simplify
\[ \sqrt{8} = 2\sqrt{2} \]
\[ \boxed{d = 2\sqrt{2}} \] Quick Tip: Always use the distance formula \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) to find the distance between two coordinate points.

Step 1: Understanding the geometry

- A tangent at point \(P\) of a circle is perpendicular to the radius \(OP\).

- Thus, \(\triangle OPQ\) is a right-angled triangle with right angle at \(P\).

- Here, \(OP = 5 \,cm\) (radius) and \(OQ = 12 \,cm\).


Step 2: Apply Pythagoras theorem
\[ OQ^2 = OP^2 + PQ^2 \]

Substitute values: \[ 12^2 = 5^2 + PQ^2 \]
\[ 144 = 25 + PQ^2 \]
\[ PQ^2 = 144 - 25 = 119 \]
\[ PQ = \sqrt{119} \,cm \]

Step 3: Conclusion

The length of \(PQ\) is \(\sqrt{119} \,cm\).


The correct answer is option (B).
Quick Tip: In circle tangent problems, always look for right-angled triangles involving radius and tangent. Pythagoras theorem is the key.

Step 1: Use similarity in the triangle

Since \(DE \parallel BC\), triangles \(\triangle ADE\) and \(\triangle ABC\) are similar. Hence the corresponding sides are proportional:
\[ \frac{DE}{BC}=\frac{AD}{AB}. \]

Step 2: Compute \(AB\) and substitute

Given \(AD=3\,cm\) and \(DB=1.5\,cm\), so
\[ AB=AD+DB=3+1.5=4.5\,cm. \]
Also \(DE=4\,cm\). Using the ratio:
\[ \frac{4}{BC}=\frac{3}{4.5}\Rightarrow BC=\frac{4\times 4.5}{3}=4\times 1.5=6\,cm. \]

Step 3: Conclusion

Therefore, the length of \(BC\) is \(6\,cm\).


The correct answer is option (D).
Quick Tip: When a line parallel to the base cuts the other two sides of a triangle, use \(\frac{segment on the smaller triangle}{corresponding side of the big triangle}=\frac{adjacent side}{whole side}\) to relate lengths quickly.

Step 1: Formula for area of a sector
\[ Area of sector = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 2: Substitute values

Here, \(\theta = 60^\circ\), \(r = 6\) cm.
\[ Area = \frac{60}{360} \times \pi \times 6^2 \] \[ = \frac{1}{6} \times \pi \times 36 = 6\pi\ cm^2 \]

Step 3: Verify carefully

Oops! Our result is \(6\pi\) cm\(^2\), not \(12\pi\). Let’s check options again:

Given options: \(36\pi, 12\pi, 6\pi, 132\). Correct match is (C) \(6\pi\ cm^2\).

\[ \boxed{Area of sector = 6\pi\ cm^2} \] Quick Tip: For a sector, always multiply \(\dfrac{\theta}{360}\) with the area of the circle \(\pi r^2\). Double-check arithmetic to avoid mistakes.

Step 1: Dimensions of the new solid

Each cube has side \(= 10\) cm. By joining two cubes face-to-face, we get a cuboid of dimensions:
\[ Length = 20\ cm, \quad Breadth = 10\ cm, \quad Height = 10\ cm \]

Step 2: Formula for surface area of a cuboid
\[ SA = 2(lb + bh + hl) \]

Step 3: Substitute values
\[ SA = 2(20 \times 10 + 10 \times 10 + 10 \times 20) \] \[ = 2(200 + 100 + 200) = 2 \times 500 = 1000\ cm^2 \]

Step 4: Re-check carefully

Wait! When two cubes are joined, the common face (area \(= 10 \times 10 = 100\)) is hidden on both cubes. So, from total surface area of two cubes:

Each cube SA \(= 6a^2 = 6 \times 100 = 600\). For two cubes: \(1200\). Subtract \(2 \times 100 = 200\) (common faces) = \(1000\).


But option (D) \(1000\) is present. So, correct is (D), not (B).

\[ \boxed{Surface Area = 1000\ cm^2} \] Quick Tip: When combining solids, remember to subtract the area of the hidden/common faces from the total surface area.

Step 1: Write quadratic polynomial in standard form
\[ 6x^2 - 7x - 3 = 0 \]
Here \(a = 6\), \(b = -7\), \(c = -3\).


Step 2: Apply quadratic formula
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-3)}}{2(6)} \]
\[ x = \frac{7 \pm \sqrt{49 + 72}}{12} \]
\[ x = \frac{7 \pm \sqrt{121}}{12} \]
\[ x = \frac{7 \pm 11}{12} \]

Step 3: Simplify roots

Case 1: \(\dfrac{7 + 11}{12} = \dfrac{18}{12} = \dfrac{3}{2}\)

Case 2: \(\dfrac{7 - 11}{12} = \dfrac{-4}{12} = -\dfrac{1}{3}\)


Step 4: Conclusion

The roots are \(\tfrac{3}{2}\) and \(-\tfrac{1}{3}\).


The correct answer is option (B).
Quick Tip: Always apply the quadratic formula carefully. Check signs of \(b\) and \(c\) to avoid errors.

Step 1: Write equations in standard form

Equation 1: \(\dfrac{4}{3}x + 2y = 8 \Rightarrow 4x + 6y = 24\)

Equation 2: \(2x + 3y = 12\)


Step 2: Compare coefficients

Equation 1: \(4x + 6y = 24\)

Equation 2: \(2x + 3y = 12\)


Divide Equation 1 by 2: \(2x + 3y = 12\)


This is exactly Equation 2.


Step 3: Interpret result

Both equations are identical, meaning they represent the same line.

Thus, there are infinitely many solutions.


Step 4: Correction

So the correct answer is (B) Infinite.
Quick Tip: For linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\): - If \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\), unique solution. - If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\), infinite solutions. - If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), no solution.

Step 1: Recall the formula for discriminant

For a quadratic \(ax^2 + bx + c = 0\): \[ D = b^2 - 4ac \]

Step 2: Identify coefficients

Here, \(a = 3\), \(b = -4\sqrt{3}\), \(c = 4\).


Step 3: Substitute values
\[ D = (-4\sqrt{3})^2 - 4(3)(4) \] \[ = 16 \times 3 - 48 \] \[ = 48 - 48 = 0 \]
\[ \boxed{D = 0} \] Quick Tip: If the discriminant \(D = 0\), the quadratic has real and equal roots. If \(D > 0\), roots are real and distinct; if \(D < 0\), roots are complex.

Step 1: Recall definition of common difference

In an arithmetic progression (A.P.), the common difference \(d\) is given by: \[ d = a_{n+1} - a_n \]

Step 2: Take consecutive terms

First term = \(3\), second term = \(3+\sqrt{2}\).
\[ d = (3+\sqrt{2}) - 3 = \sqrt{2} \]

Step 3: Verify with next terms

Third term = \(3+2\sqrt{2}\). Difference from second term: \[ (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2} \]
This matches. Hence \(d = \sqrt{2}\).

\[ \boxed{d = \sqrt{2}} \] Quick Tip: To find the common difference of an A.P., always subtract any two consecutive terms. The difference remains constant throughout.

Step 1: Assume the contrary.

Suppose \(6\sqrt{3}\) is rational. Then it can be expressed as \[ 6\sqrt{3} = \frac{m}{n}, \]
where \(m,n \in \mathbb{Z},\ n \neq 0,\ \gcd(m,n)=1\).


Step 2: Divide by the rational number 6.

Since 6 is a nonzero rational, we get \[ \sqrt{3} = \frac{m}{6n}. \]
This implies that \(\sqrt{3}\) is rational.


Step 3: Contradiction.

It is already known that \(\sqrt{3}\) is irrational. Hence, our assumption is false.


Conclusion:

Therefore, \(6\sqrt{3}\) is irrational.
Quick Tip: The product of a nonzero rational number and an irrational number is always irrational.

Step 1: Write the condition of equidistance.
\[ QP = QR \]
Using the distance formula, \[ \sqrt{(0-5)^2 + (1-(-3))^2} = \sqrt{(0-x)^2 + (1-6)^2}. \]

Step 2: Simplify both sides.

Left side: \((0-5)^2 = 25\), \((1+3)^2 = 16\) \[ QP = \sqrt{25+16} = \sqrt{41}. \]
Right side: \((0-x)^2 = x^2\), \((1-6)^2 = 25\) \[ QR = \sqrt{x^2 + 25}. \]

Step 3: Equating and solving.
\[ \sqrt{41} = \sqrt{x^2 + 25} \ \Rightarrow\ 41 = x^2 + 25 \ \Rightarrow\ x^2 = 16. \] \[ x = \pm 4. \]

Conclusion:

The values of \(x\) are \(4\) and \(-4\).
Quick Tip: For “equidistant from two points”, apply the distance formula, equate the expressions, and simplify by squaring both sides.

Let the points be \(A(5,-6)\) and \(B(-1,-4)\). Suppose the Y-axis divides \(AB\) at point \(P(0,y)\) in the ratio \(m:n\).

Using the section formula for the \(x\)-coordinate: \[ x_P = \frac{mx_2 + nx_1}{m+n} \]
Here, \(x_P = 0\), \(x_1 = 5\), \(x_2 = -1\).
\[ 0 = \frac{m(-1) + n(5)}{m+n} \] \[ 0 = \frac{-m + 5n}{m+n} \] \[ -m + 5n = 0 \quad \Rightarrow \quad m = 5n \]

Thus, the ratio is \(\; m:n = 5:1\).

Check for \(y\)-coordinate: \[ y_P = \frac{m y_2 + n y_1}{m+n} = \frac{5(-4) + 1(-6)}{6} = \frac{-20 - 6}{6} = -\frac{26}{6} = -\frac{13}{3} \]
Since \(-\frac{13}{3}\) lies between \(-6\) and \(-4\), the point lies on the segment.
\[ \boxed{The Y-axis divides the line segment in the ratio 5:1} \] Quick Tip: When finding ratios of division by an axis, always apply the section formula with \(x=0\) (for Y-axis) or \(y=0\) (for X-axis).

Starting with the Left-Hand Side (LHS): \[ \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \]

Multiply numerator and denominator by \((1-\cos\theta)\): \[ = \sqrt{\frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}} \] \[ = \sqrt{\frac{(1-\cos\theta)^2}{1-\cos^2\theta}} \] \[ = \sqrt{\frac{(1-\cos\theta)^2}{\sin^2\theta}} \] \[ = \frac{1-\cos\theta}{\sin\theta} \]

Now simplify: \[ \frac{1-\cos\theta}{\sin\theta} = \frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta} \] \[ = \csc\theta - \cot\theta \]

Hence, \[ \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \csc\theta - \cot\theta \]
\[ \boxed{Proved} \] Quick Tip: When proving trigonometric identities involving square roots, rationalize the fraction or use half-angle formulas to simplify.

Step 1: Area of sector

The angle at the centre is \(120^\circ\).

Area of sector \(OAB = \dfrac{\theta}{360^\circ} \times \pi r^2\)
\(= \dfrac{120}{360} \times \pi \times (14)^2\)
\(= \dfrac{1}{3} \times \pi \times 196\)
\(= \dfrac{196\pi}{3}\) cm\(^2\)


Step 2: Area of triangle \(OAB\)

Here, \(\triangle OAB\) is an isosceles triangle with \(OA = OB = 14\) cm and \(\angle AOB = 120^\circ\).

Area of \(\triangle OAB = \dfrac{1}{2} \times OA \times OB \times \sin(120^\circ)\)
\(= \dfrac{1}{2} \times 14 \times 14 \times \dfrac{\sqrt{3}}{2}\)
\(= 49\sqrt{3}\) cm\(^2\)


Step 3: Area of segment

Area of segment = Area of sector \(-\) Area of triangle
\(= \dfrac{196\pi}{3} - 49\sqrt{3}\) cm\(^2\)

\[ \boxed{\dfrac{196\pi}{3} - 49\sqrt{3} \ cm^2} \] Quick Tip: For finding the area of a circular segment, always subtract the area of the triangle from the area of the sector.

Step 1: Total number of outcomes

A standard deck has \(52\) cards. Hence, total outcomes \(= 52\).


(i) King of red colour

There are \(2\) red kings in the deck (hearts and diamonds).

Favourable outcomes \(= 2\).

Probability \(= \dfrac{2}{52} = \dfrac{1}{26}\).


(ii) A face card

Face cards are Jack, Queen, King of each suit.

Total face cards \(= 3 \times 4 = 12\).

Favourable outcomes \(= 12\).

Probability \(= \dfrac{12}{52} = \dfrac{3}{13}\).

\[ \boxed{(i) \ \dfrac{1}{26}, \quad (ii) \ \dfrac{3}{13}} \] Quick Tip: In a standard deck of cards, always remember: there are 52 cards, 4 suits, 13 cards in each suit, and 12 face cards in total.

Step 1: Midpoint and perpendicular from the centre.

Let \(O\) be the centre of the circle, and \(M\) the midpoint of chord \(PQ\). Then \(OM \perp PQ\).

Given \(OP = 2.5\ cm,\ PQ = 4\ cm \Rightarrow PM = 2\ cm\).

Thus, \(OM = \sqrt{OP^2 - PM^2} = \sqrt{2.5^2 - 2^2} = \sqrt{6.25 - 4} = 1.5\ cm\).


Step 2: Relation between polar and pole.

Since \(T\) is the intersection of tangents at \(P\) and \(Q\), chord \(PQ\) is the polar of \(T\).
For a circle, if the perpendicular distance from \(O\) to the polar is \(d\), and \(OT\) is the distance of the pole from \(O\), then: \[ d = \frac{r^2}{OT}. \]
Here \(d=1.5,\ r=2.5 \Rightarrow OT=\frac{2.5^2}{1.5}=\frac{6.25}{1.5}=\frac{25}{6}\ cm. \] \textbf{Step 3: Tangent length.}
By power of a point: \)TP^2 = OT^2 - r^2\(.
\[ TP = \sqrt{\left(\frac{25}{6}\right)^2 - (2.5)^2} = \sqrt{\frac{625}{36} - \frac{25}{4}} = \sqrt{\frac{400}{36}} = \frac{10}{3}\ cm. \]
\[ \boxed{TP = \tfrac{10}{3}\ cm} \] Quick Tip: For circle tangent problems, combine the perpendicular from the centre to a chord with the pole–polar relation and apply power of a point.

Step 1: Apply BPT in \(\triangle ABC\).

Since \(DE \parallel AC\), by Basic Proportionality Theorem (BPT): \[ \frac{BD}{BA} = \frac{BE}{BC}. \quad (1) \]

Step 2: Apply BPT in \(\triangle ABE\).

Since \(DF \parallel AE\): \[ \frac{BD}{BA} = \frac{BF}{BE}. \quad (2) \]

Step 3: Equating ratios.

From (1) and (2): \[ \frac{BF}{BE} = \frac{BE}{BC} \quad \Rightarrow \quad BF = \frac{BE^2}{BC}. \]

Step 4: Express \(FE\).
\(FE = BE - BF = BE - \frac{BE^2}{BC} = BE\left(1 - \frac{BE}{BC}\right) = BE \cdot \frac{BC - BE}{BC} = BE \cdot \frac{EC}{BC}\).

Step 5: Required ratio.
\[ \frac{BF}{FE} = \frac{\tfrac{BE^2}{BC}}{\tfrac{BE \cdot EC}{BC}} = \frac{BE}{EC}. \]
\[ \boxed{\dfrac{BF}{FE} = \dfrac{BE}{EC}} \] Quick Tip: When parallels are drawn in triangles, always apply the Basic Proportionality Theorem (BPT) in two different triangles and equate the common ratios to establish the required relation.

Step 1: Write the sum formula of an A.P.

For an A.P. with first term \(a\) and common difference \(d\), \[ S_n=\frac{n}{2}\big(2a+(n-1)d\big). \]

Step 2: Form equations from given sums.

For \(n=9\): \[ S_9=\frac{9}{2}(2a+8d)=81 \Rightarrow 9(2a+8d)=162 \Rightarrow 2a+8d=18 \Rightarrow a+4d=9.\tag{1} \]
For \(n=17\): \[ S_{17}=\frac{17}{2}(2a+16d)=289 \Rightarrow 17(2a+16d)=578 \Rightarrow 2a+16d=34 \Rightarrow a+8d=17.\tag{2} \]

Step 3: Solve (1) and (2).

Subtract (1) from (2): \((a+8d)-(a+4d)=17-9 \Rightarrow 4d=8 \Rightarrow d=2\).

Substitute in (1): \(a+4(2)=9 \Rightarrow a+8=9 \Rightarrow a=1\).

\boxed{a=1,\quad d=2. Quick Tip: When two different partial sums of an A.P. are given, convert each to a linear equation in \(a\) and \(d\) using \(S_n=\frac{n}{2}(2a+(n-1)d)\), then solve the pair.

Step 1: Find the zeroes.

Solve \(x^2-5=0 \Rightarrow x=\pm\sqrt{5}\). Thus the zeroes are \(\alpha=\sqrt{5}\) and \(\beta=-\sqrt{5}\).

Step 2: Verify relationships.

For \(ax^2+bx+c\), \emph{sum of zeroes \(=\,-\dfrac{b}{a}\) and \emph{product \(=\,\dfrac{c}{a}\).

Here \(a=1,\ b=0,\ c=-5\).

Sum: \(\alpha+\beta=\sqrt{5}+(-\sqrt{5})=0\ = -\dfrac{b}{a}=-\dfrac{0}{1}=0\).

Product: \(\alpha\beta=\sqrt{5}\cdot(-\sqrt{5})=-5\ = \dfrac{c}{a}=\dfrac{-5}{1}=-5\).

\boxed{\text{Zeroes: \sqrt{5,\ -\sqrt{5\ \ \text{and both relations are verified. Quick Tip: For \(x^2+bx+c\), the zeroes’ sum is \(-b\) and product is \(c\). Spot \(b\) and \(c\) quickly to verify without fully solving.

Class mark (\(x_i\)): \(50,60,70,80,90\) with frequencies (\(f_i\)): \(3,10,11,8,3\).

Compute \(\sum f_i x_i\): \(3(50)+10(60)+11(70)+8(80)+3(90)=150+600+770+640+270=2430\).

Total \(N=\sum f_i=35\).
\[ Mean=\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2430}{35}=69.428571\ldots \approx \boxed{69.43%} \] Quick Tip: For grouped data with equal class width, use class marks \(x_i\) and the formula \(\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i}\). A quick check: the mean should lie near the heaviest frequencies (here around \(65\)--\(75\)), so \(\approx 69%\) makes sense.

Total \(N=6+11+21+23+14+5=80\), class width \(h=10\).

Mode:

Modal class \(=35\)--\(45\) (highest frequency \(f_1=23\)). With \(l=35,\ f_0=21,\ f_2=14\),
\[ Mode=l+\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}\,h =35+\frac{23-21}{2\cdot23-21-14}\times10 =35+\frac{2}{11}\times10 =35+\frac{20}{11}\approx \boxed{36.82\ years}. \]

Median:

Cumulative frequencies: \(6,17,38,61,75,80\). Since \(N/2=40\), the median class is \(35\)--\(45\) (\(l=35\)) with \(f=23\) and cumulative before it \(c_f=38\).
\[ Median=l+\frac{\left(\frac{N}{2}-c_f\right)}{f}\,h =35+\frac{(40-38)}{23}\times10 =35+\frac{20}{23} \approx \boxed{35.87\ years}. \] Quick Tip: For the mode in grouped data, use \(l+\dfrac{(f_1-f_0)}{(2f_1-f_0-f_2)}h\). For the median, locate the class where the cumulative frequency first exceeds \(N/2\), then apply \(l+\dfrac{(N/2-c_f)}{f}h\).

N/A Quick Tip: When a two-digit number and its reverse are involved, set the number as \(10a+b\); reversing swaps \(a\) and \(b\). The equation \(11(a+b)\) often appears from adding them.

Let the width of the rectangular field be \(x\) metres.

Then, length of the field \(= 2x + 9\) metres.


Step 1: Write the area equation
\[ Area = Length \times Width \] \[ 810 = (2x+9)\cdot x \]

Step 2: Form the quadratic equation
\[ 810 = 2x^2 + 9x \] \[ 2x^2 + 9x - 810 = 0 \]

Step 3: Simplify

Divide throughout by 2: \[ x^2 + \tfrac{9}{2}x - 405 = 0 \]
Multiply through by 2 to avoid fraction: \[ 2x^2 + 9x - 810 = 0 \]

Step 4: Solve using quadratic formula
\[ x = \frac{-9 \pm \sqrt{9^2 - 4(2)(-810)}}{2(2)} \] \[ x = \frac{-9 \pm \sqrt{81 + 6480}}{4} \] \[ x = \frac{-9 \pm \sqrt{6561}}{4} \] \[ x = \frac{-9 \pm 81}{4} \]

So, \[ x = \frac{72}{4}=18 \quad or \quad x = \frac{-90}{4}=-22.5 \]
Width cannot be negative, so \(x=18\).


Step 5: Find length
\[ Length = 2x + 9 = 2(18)+9 = 45 \]
\[ \boxed{Width = 18 m, Length = 45 m} \] Quick Tip: In word problems, always form an equation step-by-step. For rectangular fields, width is often taken as \(x\), and length is expressed in terms of \(x\). Then apply area = length × width.

Let the height of the tower be \(h\). For an altitude angle \(\alpha\), the shadow length is \(h\cot\alpha\).

Shadow at \(30^\circ\): \(h\cot30^\circ=h\sqrt{3}\).

Shadow at \(60^\circ\): \(h\cot60^\circ=\dfrac{h}{\sqrt{3}}\).

Given: \(h\sqrt{3}-\dfrac{h}{\sqrt{3}}=30 \Rightarrow h\left(\dfrac{3-1}{\sqrt{3}}\right)=30 \Rightarrow h\cdot\dfrac{2}{\sqrt{3}}=30\).
\(\Rightarrow\ h=15\sqrt{3}\ m=15(1.732)\ m=25.98\ m\ (\approx 26\ m).\)

\[ \boxed{h=15\sqrt{3}\ m\ \approx\ 26\ m} \] Quick Tip: For sun–shadow problems, use \(\,shadow=h\cot\alpha\,\) and the given difference to form a linear equation in \(h\).

Using half–angle identities: \(1-\cos\theta=2\sin^2\frac{\theta}{2}\) and \(\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\). Put \(s=\sin\frac{\theta}{2}\) and \(c=\cos\frac{\theta}{2}\). Then
\[ \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} =\frac{2sc-(c^2-s^2)+1}{2sc+(c^2-s^2)-1} =\frac{2sc+2s^2}{2sc-2s^2} =\frac{c+s}{\,c-s\,}. \tag{1} \]
Now, \[ \sec\theta+\tan\theta=\frac{1+\sin\theta}{\cos\theta} =\frac{1+2sc}{c^2-s^2} =\frac{(s+c)^2}{(c+s)(c-s)} =\frac{c+s}{\,c-s\,}. \tag{2} \]
From (1) and (2), both sides are equal; hence proved.

\[ \boxed{\displaystyle \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\sec\theta+\tan\theta} \] Quick Tip: When RHS contains \(\sec\theta+\tan\theta\), try rewriting it as \(\dfrac{1+\sin\theta}{\cos\theta}\) or use half–angle substitutions \(s=\sin\frac{\theta}{2},\,c=\cos\frac{\theta}{2}\) to simplify the LHS.

Step 1: Greatest possible diameter.

The hemisphere is placed on the \emph{top face of the cube (a square of side \(14\) cm). For it to fit exactly, its circular base must be inscribed in the square. Hence the largest possible diameter equals the side of the square: \[ \boxed{Greatest diameter =14 cm},\qquad r=\frac{14}{2}=7 cm. \]

Step 2: Exposed surface area.

Exposed parts: (i) Curved surface area (CSA) of the hemisphere \(=2\pi r^2\), (ii) Five faces of the cube (all except the top, which is covered).

So, \[ S=2\pi r^2+5\cdot(14)^2 =2\cdot 3.14\cdot 7^2+5\cdot 196 =2\cdot 3.14\cdot 49+980 =307.72+980. \] \[ \boxed{S=1287.72\ cm^2}. \] Quick Tip: When a hemisphere sits perfectly on a cube, the top face is fully covered. Add the hemisphere’s CSA to only \emph{five} faces of the cube.

Step 1: Find the cone’s height.

Given \(r=2\) cm and \(l=2\sqrt{2}\) cm, so \[ h=\sqrt{l^2-r^2}=\sqrt{(2\sqrt{2})^2-2^2}=\sqrt{8-4}=2 cm. \]

Step 2: Volumes.

Cone: \(V_{cone}=\dfrac{1}{3}\pi r^2 h=\dfrac{1}{3}\pi\cdot 4\cdot 2=\dfrac{8}{3}\pi\).

Hemisphere: \(V_{hem}=\dfrac{2}{3}\pi r^3=\dfrac{2}{3}\pi\cdot 8=\dfrac{16}{3}\pi\).

Total: \[ V=V_{cone}+V_{hem}=\frac{8}{3}\pi+\frac{16}{3}\pi=\frac{24}{3}\pi=8\pi. \]
With \(\pi=3.14\), \[ \boxed{V=8\pi=25.12\ cm^3}. \] Quick Tip: For a cone with given slant height \(l\) and radius \(r\), compute \(h=\sqrt{l^2-r^2}\) first; then add volumes part-wise when solids are combined.