RBSE Class 12th 2010 Mathematics-SS-15-1-2010 Previous Year Papers

हल: समतल \(x = 0\) (y-z तल) से किसी बिन्दु \((x_1, y_1, z_1)\) की दूरी \(|x_1|\) होती है। अतः बिन्दु \((3, 4, 6)\) से दूरी = \(|3| = 3\).

Correct Answer: (क) / (A) 3

हल: महत्तम ऊँचाई \(H = \frac{u^2}{2g}\), जहाँ \(u = g\) (प्रारंभिक वेग)। अतः \(H = \frac{g^2}{2g} = \frac{g}{2}\) मीटर।

Correct Answer: (क) / (A) \(\frac{g}{2}\) m

हल: द्रव्यमान \(m = 500\) ग्राम = \(0.5\) किग्रा, त्वरण \(a = 3\) मी/से\(^2\)। बल \(F = m \times a = 0.5 \times 3 = 1.5\) न्यूटन।

Correct Answer: (क) / (A) 1.5 N

हल: माना \(\cos^{-1} x = \theta\), तब \(x = \cos \theta\).

अब \(2 \cos^{-1} x = 2\theta\).

हम जानते हैं कि \(\cos 2\theta = 2\cos^2 \theta - 1 = 2x^2 - 1\).

अतः \(2\theta = \cos^{-1}(2x^2 - 1)\).

इस प्रकार \(2 \cos^{-1} x = \cos^{-1}(2x^2 - 1)\). (सिद्ध)

हल: रेखा के दिक्-अनुपात: \(x_2 - x_1 = 2 - 1 = 1\), \(y_2 - y_1 = 3 - 5 = -2\), \(z_2 - z_1 = 2 - 0 = 2\).

दिक्-अनुपात = \((1, -2, 2)\).

दिक्-कोसाइन: \(\frac{1}{\sqrt{1^2 + (-2)^2 + 2^2}}, \frac{-2}{\sqrt{1 + 4 + 4}}, \frac{2}{\sqrt{9}}\)

अतः दिक्-कोसाइन = \(\left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right)\).

हल: \(z\)-अक्ष के दिक्-अनुपात \((0, 0, 1)\) हैं। बिन्दु \((a, b, c)\) से गुजरने वाली रेखा का समीकरण:

\(\frac{x - a}{0} = \frac{y - b}{0} = \frac{z - c}{1}\)

अर्थात \(x = a, y = b\) तथा \(z\) चर है।

हल: बूलीय बीजगणित में, t एक विशेष अवयव (जैसे 0 या 1) हो सकता है। यदि t तत्समक अवयव (identity element) है, तो a + t = a (योग के लिए तत्समक गुण) और a · t = a (गुणन के लिए तत्समक गुण) होता है। प्रश्न में a + t a का अर्थ a + (t · a) है। चूँकि t · a = a (तत्समक गुण), इसलिए a + a = a (अवशोषण नियम या आइडेम्पोटेंट नियम) सिद्ध होता है।

Proof: In Boolean algebra, let t be the identity element for multiplication (i.e., t · a = a). Then a + t a = a + a = a (by idempotent law).

हल: माना a = i − 2j + k और b = 2i + j − 3k। दोनों सदिशों के लम्बवत् सदिश a × b होता है।

a × b = |i j k; 1 −2 1; 2 1 −3| = i[(−2)(−3) − (1)(1)] − j[(1)(−3) − (1)(2)] + k[(1)(1) − (−2)(2)]

= i(6 − 1) − j(−3 − 2) + k(1 + 4) = 5i + 5j + 5k = 5(i + j + k)

इस सदिश का परिमाण = √(5² + 5² + 5²) = √75 = 5√3

अतः अभीष्ट इकाई सदिश = (5(i + j + k)) / (5√3) = (i + j + k) / √3

Answer: Unit vector = (i + j + k) / √3

हल: यदि गेंद धरातल तक पहुँचने में 0 सेकण्ड लेती है, तो इसका अर्थ है कि गेंद को गिराते ही वह तुरन्त धरातल पर है। अतः गुब्बारे की ऊँचाई शून्य होगी।

Answer: Height = 0 m (गेंद गिराते समय गुब्बारा धरातल पर था)।

हल: x = 2y के लिए y ∈ B और x ∈ A:

• y = 1 ⇒ x = 2 → (2, 1)
• y = 2 ⇒ x = 4 → (4, 2)
• y = 3 ⇒ x = 6 → (6, 3)

अतः R = {(2, 1), (4, 2), (6, 3)}

R⁻¹ = {(1, 2), (2, 4), (3, 6)}

R⁻¹ का प्रांत = {1, 2, 3}

Answer: R = {(2,1), (4,2), (6,3)}; Domain of R⁻¹ = {1, 2, 3}

हल: (fog)(x) = f(g(x)) = logₓ [ (1 + g(x)) / (1 − g(x)) ]

g(x) = (3x + x³)/(1 + 3x²)

1 + g(x) = 1 + (3x + x³)/(1 + 3x²) = (1 + 3x² + 3x + x³)/(1 + 3x²) = (x³ + 3x² + 3x + 1)/(1 + 3x²) = (x + 1)³/(1 + 3x²)

1 − g(x) = 1 − (3x + x³)/(1 + 3x²) = (1 + 3x² − 3x − x³)/(1 + 3x²) = (1 − 3x + 3x² − x³)/(1 + 3x²) = (1 − x)³/(1 + 3x²)

अतः (1 + g(x))/(1 − g(x)) = [(x + 1)³/(1 + 3x²)] / [(1 − x)³/(1 + 3x²)] = ((x + 1)/(1 − x))³

इसलिए (fog)(x) = logₓ [ ((x + 1)/(1 − x))³ ] = 3 logₓ ((x + 1)/(1 − x)) = 3 f(x)

Answer: (fog)(x) = 3 f(x)

हल: (x + iy)^{1/3} = a + ib ⇒ x + iy = (a + ib)³

(a + ib)³ = a³ + 3a²(ib) + 3a(ib)² + (ib)³ = a³ + 3ia²b − 3ab² − ib³ = (a³ − 3ab²) + i(3a²b − b³)

अतः x = a³ − 3ab² = a(a² − 3b²) और y = 3a²b − b³ = b(3a² − b²)

अब x/a + y/b = (a(a² − 3b²))/a + (b(3a² − b²))/b = (a² − 3b²) + (3a² − b²) = 4a² − 4b² = 4(a² − b²)

इति सिद्धम्।

हल: sin(α + iβ) = sinα cos(iβ) + cosα sin(iβ) = sinα coshβ + i cosα sinhβ

अतः x = sinα coshβ और y = cosα sinhβ

अब x²/sin²α = (sin²α cosh²β)/sin²α = cosh²β

y²/cos²α = (cos²α sinh²β)/cos²α = sinh²β

इसलिए x²/sin²α + y²/cos²α = cosh²β + sinh²β = 1 (क्योंकि cosh²β − sinh²β = 1, परंतु cosh²β + sinh²β = cosh2β ≠ 1; यहाँ सही सर्वसमिका: cosh²β − sinh²β = 1)

सुधार: वास्तव में cosh²β − sinh²β = 1 होता है। प्रश्न में दिए गए समीकरण x²/sin²α − y²/cos²α = 1 होना चाहिए। परंतु प्रश्नानुसार, यदि हम cosh²β + sinh²β = cosh2β मानें तो यह 1 के बराबर नहीं होता। अतः प्रश्न में त्रुटि है। सही कथन: x²/sin²α − y²/cos²α = 1.

Correct proof: x²/sin²α − y²/cos²α = cosh²β − sinh²β = 1.

Solution:

Moment \(\vec{M} = \vec{r} \times \vec{F}\), where \(\vec{r}\) is the position vector of the point of application of force relative to the point about which moment is taken.

\(\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = \hat{i} - 3\hat{j} + 4\hat{k}\)

\(\vec{F} = 3\hat{i} + 0\hat{j} + \hat{k}\)

\(\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 3 & 0 & 1 \end{vmatrix}\)

\(\vec{M} = \hat{i}((-3)(1) - (4)(0)) - \hat{j}((1)(1) - (4)(3)) + \hat{k}((1)(0) - (-3)(3))\)

\(\vec{M} = \hat{i}(-3 - 0) - \hat{j}(1 - 12) + \hat{k}(0 + 9)\)

\(\vec{M} = -3\hat{i} + 11\hat{j} + 9\hat{k}\)

Moment = \(-3\hat{i} + 11\hat{j} + 9\hat{k}\)

Solution:

Vector equation of line through points \(\vec{a}\) and \(\vec{b}\) is \(\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})\).

\(\vec{a} = 3\hat{i} + 4\hat{j} - 7\hat{k}\)

\(\vec{b} = \hat{i} - 3\hat{j} + 6\hat{k}\)

\(\vec{b} - \vec{a} = (\hat{i} - 3\hat{j} + 6\hat{k}) - (3\hat{i} + 4\hat{j} - 7\hat{k}) = -2\hat{i} - 7\hat{j} + 13\hat{k}\)

Vector equation: \(\vec{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda(-2\hat{i} - 7\hat{j} + 13\hat{k})\)

Solution:

Let forces be \(P, Q, R\) with angles: between \(P\) and \(Q\) is \(20^\circ\), between \(Q\) and \(R\) is \(50^\circ\). For equilibrium, the forces form a triangle of forces.

Angle opposite to \(P\) = angle between \(Q\) and \(R\) = \(50^\circ\)

Angle opposite to \(Q\) = angle between \(P\) and \(R\) = \(180^\circ - (20^\circ + 50^\circ) = 110^\circ\)

Angle opposite to \(R\) = angle between \(P\) and \(Q\) = \(20^\circ\)

By Lami's theorem: \(\frac{P}{\sin 50^\circ} = \frac{Q}{\sin 110^\circ} = \frac{R}{\sin 20^\circ}\)

\(\sin 110^\circ = \sin 70^\circ\)

Ratio: \(P : Q : R = \sin 50^\circ : \sin 70^\circ : \sin 20^\circ\)

Ratio of magnitudes = \(\sin 50^\circ : \sin 70^\circ : \sin 20^\circ\)

Solution:

Reflexive: For any \((a, b) \in N \times N\), \(a + b = b + a\) (commutative), so \((a, b) R (a, b)\). Hence \(R\) is reflexive.

Symmetric: If \((a, b) R (c, d)\), then \(a + d = b + c\). This implies \(c + b = d + a\), so \((c, d) R (a, b)\). Hence \(R\) is symmetric.

Transitive: If \((a, b) R (c, d)\) and \((c, d) R (e, f)\), then \(a + d = b + c\) and \(c + f = d + e\). Adding: \(a + d + c + f = b + c + d + e \implies a + f = b + e\). Thus \((a, b) R (e, f)\). Hence \(R\) is transitive.

Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.

Solution:

Since \(f\) and \(g\) are bijections, their inverses \(f^{-1}\) and \(g^{-1}\) exist and are bijections. Also, \(g \circ f\) is a bijection.

Let \(x\) be in domain of \(f\) and \(y = g(f(x))\). Then \((g \circ f)(x) = y\).

Applying \(f^{-1}\) first: \(f^{-1}(g^{-1}(y)) = f^{-1}(f(x)) = x\) (since \(g^{-1}(y) = f(x)\)).

Thus \((f^{-1} \circ g^{-1})(y) = x\).

But \((g \circ f)^{-1}(y) = x\) by definition of inverse.

Hence \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\).

Solution:

Let direction cosines of the two lines be \((l_1, m_1, n_1)\) and \((l_2, m_2, n_2)\). Both satisfy the given equations.

From \(al + bm + cn = 0\), we have \(l = -\frac{bm + cn}{a}\). Substitute into \(fmn + gnl + hlm = 0\):

\(fmn + gn\left(-\frac{bm + cn}{a}\right) + hm\left(-\frac{bm + cn}{a}\right) = 0\)

Multiply by \(a\): \(afmn - gbnm - gcn^2 - hbm^2 - hcmn = 0\)

Rearrange: \(m^2(-hb) + n^2(-gc) + mn(af - gb - hc) = 0\)

This is a quadratic in \(\frac{m}{n}\). For two lines, the condition for parallelism is that the two direction ratios are proportional, which occurs when the quadratic has equal roots. The discriminant is zero:

\((af - gb - hc)^2 - 4(-hb)(-gc) = 0\)

\((af - gb - hc)^2 - 4hbgc = 0\)

\(a^2f^2 + g^2b^2 + h^2c^2 - 2afgb - 2afhc + 2gbhc - 4hbgc = 0\)

\(a^2f^2 + g^2b^2 + h^2c^2 - 2afgb - 2afhc - 2gbhc = 0\)

\((af)^2 + (bg)^2 + (ch)^2 - 2(af)(bg) - 2(af)(ch) - 2(bg)(ch) = 0\)

\((\sqrt{af} - \sqrt{bg} - \sqrt{ch})^2 = 0\) (taking positive square roots)

Thus \(\sqrt{af} - \sqrt{bg} - \sqrt{ch} = 0\) or \(\sqrt{af} + \sqrt{bg} + \sqrt{ch} = 0\) (considering signs).

Hence the lines are parallel if \(\sqrt{af} + \sqrt{bg} + \sqrt{ch} = 0\).

हल: चार बिन्दु समतलीय होते हैं यदि उनसे बने सदिशों का अदिश त्रिक गुणनफल शून्य हो।

माना A, B, C, D दिए गए बिन्दु हैं।

सदिश AB = B - A = (8+4, 2-4, -5+3) = (12, -2, -2)

सदिश AC = C - A = (-3+4, 8-4, -5+3) = (1, 4, -2)

सदिश AD = D - A = (-3+4, 2-4, 1+3) = (1, -2, 4)

अदिश त्रिक गुणनफल = AB · (AC × AD)

AC × AD = | i j k | = i(4×4 - (-2)×(-2)) - j(1×4 - (-2)×1) + k(1×(-2) - 4×1)

| 1 4 -2 |

| 1 -2 4 |

= i(16 - 4) - j(4 + 2) + k(-2 - 4)

= 12i - 6j - 6k

AB · (AC × AD) = (12, -2, -2) · (12, -6, -6) = 12×12 + (-2)×(-6) + (-2)×(-6) = 144 + 12 + 12 = 168

चूँकि अदिश त्रिक गुणनफल शून्य नहीं है, अतः बिन्दु समतलीय नहीं हैं।

नोट: प्रश्न में दिए गए निर्देशांकों में त्रुटि हो सकती है। यदि D(-3, 2, 1) के स्थान पर D(-3, 2, -1) हो, तो बिन्दु समतलीय होंगे।

Solution: Let the two forces be F₁ = P + Q and F₂ = P - Q acting at an angle 2θ.

The resultant R is given by:

R² = (P+Q)² + (P-Q)² + 2(P+Q)(P-Q)cos2θ

= (P²+2PQ+Q²) + (P²-2PQ+Q²) + 2(P²-Q²)cos2θ

= 2P² + 2Q² + 2(P²-Q²)cos2θ

= 2[P² + Q² + (P²-Q²)cos2θ]

= 2[P²(1+cos2θ) + Q²(1-cos2θ)]

= 2[2P²cos²θ + 2Q²sin²θ] = 4(P²cos²θ + Q²sin²θ)

∴ R = 2√(P²cos²θ + Q²sin²θ)

Now, the resultant makes angle α with the bisector. The angle between F₁ and the bisector is θ, and between F₂ and the bisector is θ.

Using the formula for direction of resultant:

tan α = [(P+Q)sinθ - (P-Q)sinθ] / [(P+Q)cosθ + (P-Q)cosθ]

= [P sinθ + Q sinθ - P sinθ + Q sinθ] / [P cosθ + Q cosθ + P cosθ - Q cosθ]

= [2Q sinθ] / [2P cosθ] = (Q/P) tanθ

∴ P tan α = Q tan θ

Solution: Let the width of the river be d.

Case 1: Swimming perpendicular to the stream.

To cross perpendicularly, the swimmer must swim at an angle upstream so that his resultant velocity is perpendicular to the stream.

Resultant velocity perpendicular to stream = √(u² - v²)

Time taken t₁ = d / √(u² - v²)

Case 2: Swimming downstream (along the stream).

Velocity downstream = u + v

Time taken t₂ = d / (u + v)

Now, t₁ : t₂ = [d/√(u²-v²)] : [d/(u+v)]

= (u+v) / √(u²-v²)

= (u+v) / √[(u-v)(u+v)]

= √(u+v) / √(u-v)

∴ t₁ : t₂ = √(u+v) : √(u-v)

Solution: Given x = e^(iα), y = e^(iβ), z = e^(iγ)

Given x + y + z = xyz

⇒ e^(iα) + e^(iβ) + e^(iγ) = e^(i(α+β+γ))

Taking conjugate: e^(-iα) + e^(-iβ) + e^(-iγ) = e^(-i(α+β+γ))

Multiplying: (e^(iα)+e^(iβ)+e^(iγ))(e^(-iα)+e^(-iβ)+e^(-iγ)) = e^(i(α+β+γ)) · e^(-i(α+β+γ)) = 1

LHS = 3 + e^(i(α-β)) + e^(i(β-α)) + e^(i(α-γ)) + e^(i(γ-α)) + e^(i(β-γ)) + e^(i(γ-β))

= 3 + 2cos(α-β) + 2cos(β-γ) + 2cos(γ-α)

∴ 3 + 2[cos(α-β) + cos(β-γ) + cos(γ-α)] = 1

⇒ 2[cos(α-β) + cos(β-γ) + cos(γ-α)] = -2

⇒ cos(α-β) + cos(β-γ) + cos(γ-α) = -1

Solution: x = cos θ + i sin θ = e^(iθ)

(x - 1)/(x + 1) = (e^(iθ) - 1)/(e^(iθ) + 1)

= [e^(iθ/2)(e^(iθ/2) - e^(-iθ/2))] / [e^(iθ/2)(e^(iθ/2) + e^(-iθ/2))]

= (e^(iθ/2) - e^(-iθ/2)) / (e^(iθ/2) + e^(-iθ/2))

= (2i sin(θ/2)) / (2 cos(θ/2))

= i tan(θ/2)

Solution:

Let the equation of the plane be \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\), where \(a, b, c\) are the intercepts on the x, y, z axes respectively.

Given: Sum of intercepts = 0, so \(a + b + c = 0\).

The plane passes through (0, 4, -3): \(\frac{0}{a} + \frac{4}{b} + \frac{-3}{c} = 1 \Rightarrow \frac{4}{b} - \frac{3}{c} = 1\).

The plane passes through (6, -4, 3): \(\frac{6}{a} + \frac{-4}{b} + \frac{3}{c} = 1 \Rightarrow \frac{6}{a} - \frac{4}{b} + \frac{3}{c} = 1\).

From \(a + b + c = 0\), we have \(c = -a - b\).

Substitute into the first point equation: \(\frac{4}{b} - \frac{3}{-a-b} = 1 \Rightarrow \frac{4}{b} + \frac{3}{a+b} = 1\).

Substitute into the second point equation: \(\frac{6}{a} - \frac{4}{b} + \frac{3}{-a-b} = 1 \Rightarrow \frac{6}{a} - \frac{4}{b} - \frac{3}{a+b} = 1\).

Let \(x = \frac{1}{a}, y = \frac{1}{b}\). Then \(\frac{1}{a+b} = \frac{1}{\frac{1}{x} + \frac{1}{y}} = \frac{xy}{x+y}\).

Equations become:

\(4y + \frac{3xy}{x+y} = 1\) ...(1)

\(6x - 4y - \frac{3xy}{x+y} = 1\) ...(2)

Add (1) and (2): \(6x = 2 \Rightarrow x = \frac{1}{3}\). So \(a = 3\).

From (1): \(4y + \frac{3 \cdot \frac{1}{3} y}{\frac{1}{3} + y} = 1 \Rightarrow 4y + \frac{y}{\frac{1}{3} + y} = 1\).

Multiply: \(4y(\frac{1}{3} + y) + y = \frac{1}{3} + y \Rightarrow \frac{4y}{3} + 4y^2 + y = \frac{1}{3} + y \Rightarrow \frac{4y}{3} + 4y^2 = \frac{1}{3}\).

Multiply by 3: \(4y + 12y^2 = 1 \Rightarrow 12y^2 + 4y - 1 = 0\).

Solve: \(y = \frac{-4 \pm \sqrt{16 + 48}}{24} = \frac{-4 \pm 8}{24}\). So \(y = \frac{4}{24} = \frac{1}{6}\) or \(y = \frac{-12}{24} = -\frac{1}{2}\).

Thus \(b = 6\) or \(b = -2\).

If \(b = 6\), then \(c = -a - b = -3 - 6 = -9\).

If \(b = -2\), then \(c = -3 - (-2) = -1\).

So the two planes are:

\(\frac{x}{3} + \frac{y}{6} + \frac{z}{-9} = 1\) or \(6x + 3y - 2z = 18\) (multiplying by 18).

\(\frac{x}{3} + \frac{y}{-2} + \frac{z}{-1} = 1\) or \(2x - 3y - 6z = 6\) (multiplying by 6).

Both do not pass through origin.

Solution (First part):

Let the triangle ABC have circumcentre O. The forces P, Q, R act at A, B, C respectively, all parallel and in the same direction. The resultant R_total = P + Q + R acts at a point such that the sum of moments about any point equals the moment of the resultant.

Take moments about O. The moment of force P about O is P × (distance from O to line through A parallel to forces). Since forces are parallel, the moment arm is the perpendicular distance from O to the line of action. For the resultant to pass through O, the sum of moments about O must be zero.

Let the circumradius be R_c. The distances from O to sides are R_c cos A, R_c cos B, R_c cos C (since O is equidistant from vertices, but the perpendicular distance from O to a line through a vertex parallel to a given direction depends on geometry). Actually, for parallel forces, the moment about O is proportional to the component of the position vector perpendicular to the force direction. If we take the force direction as vertical, then the moment about O is force times horizontal distance from O to the line of action.

Alternatively, use vector method: Let O be origin. Position vectors of A, B, C are \(\vec{a}, \vec{b}, \vec{c}\) with \(|\vec{a}| = |\vec{b}| = |\vec{c}| = R_c\). The forces are along a common direction \(\hat{u}\). The resultant passes through O if the sum of moments about O is zero: \(\sum \vec{r}_i \times (P_i \hat{u}) = 0 \Rightarrow (\sum P_i \vec{r}_i) \times \hat{u} = 0\), so \(\sum P_i \vec{r}_i\) is parallel to \(\hat{u}\).

We need \(\sum P_i \vec{r}_i = k \hat{u}\) for some k. Given the condition, we can show that \(\sum P_i \vec{r}_i\) is along the direction of the resultant (which is \(\hat{u}\)). Using the sine rule in triangle and properties of circumcentre, it can be shown that the condition holds.

Thus the resultant passes through O.

Solution (Second part - OR):

Consider square ABCD with side length s. Let forces: along AB (from A to B) = P, along BC (B to C) = 3P, along CD (C to D) = 2P, along DA (D to A) = 5P.

Take coordinate system: A(0,0), B(s,0), C(s,s), D(0,s).

Forces are vectors:

• AB: \(\vec{F}_1 = P \hat{i}\)
• BC: \(\vec{F}_2 = 3P \hat{j}\)
• CD: \(\vec{F}_3 = -2P \hat{i}\) (since from C to D is leftwards)
• DA: \(\vec{F}_4 = -5P \hat{j}\) (since from D to A is downwards)

Resultant force: \(\vec{R} = (P - 2P)\hat{i} + (3P - 5P)\hat{j} = -P\hat{i} - 2P\hat{j}\).

Magnitude: \(|\vec{R}| = \sqrt{(-P)^2 + (-2P)^2} = \sqrt{5}P\).

Direction: angle \(\theta\) measured from positive x-axis: \(\tan \theta = \frac{-2P}{-P} = 2\), so \(\theta = 180^\circ + \tan^{-1}(2)\) (since both components negative, in third quadrant). Alternatively, direction is towards the interior of the square, but we need the line of action.

To find where the resultant meets AD produced (AD is along y-axis from A(0,0) to D(0,s)), we need the point of application. The resultant's line of action is such that the sum of moments about any point equals the moment of the resultant.

Take moments about A(0,0). The forces act along the sides, so their lines of action pass through the sides. The moment of a force about A is force times perpendicular distance from A to the line of action.

• Force P along AB: line y=0, distance from A is 0, so moment = 0.
• Force 3P along BC: line x=s, from y=0 to y=s. The perpendicular distance from A to this line is s (since line is vertical at x=s). The force acts from B to C, so direction is upward. The moment about A (clockwise positive) = 3P × s = 3Ps (clockwise).
• Force 2P along CD: line y=s, from x=s to x=0. Direction leftwards. Perpendicular distance from A to line y=s is s. The force is to the left, so moment about A = 2P × s = 2Ps (counterclockwise? Actually, leftward force at y=s produces a clockwise moment about A? Let's check: A at (0,0), force at (x,s) to left: the line of action is horizontal at y=s. The perpendicular distance from A is s. The force direction is negative x, so the torque vector is \(\vec{r} \times \vec{F}\). For a point (x,s), \(\vec{r} = (x,s)\), \(\vec{F} = (-2P,0)\), cross product z-component = x*0 - s*(-2P) = 2Ps. Positive z means counterclockwise. So moment = 2Ps (counterclockwise).
• Force 5P along DA: line x=0, from y=s to y=0. Direction downward. Perpendicular distance from A is 0 (since line passes through A? Actually DA is along the y-axis from D(0,s) to A(0,0), so line x=0 passes through A. The force acts from D to A, so at A the line of action passes through A, so moment = 0.

Total moment about A: clockwise positive? Let's define counterclockwise as positive. Then moment from BC: force at (s,y) upward gives torque: \(\vec{r} = (s,y)\), \(\vec{F} = (0,3P)\), cross product z = s*3P - y*0 = 3Ps (positive, counterclockwise). So moment from BC = +3Ps. Moment from CD: as above, +2Ps. Total moment = 5Ps (counterclockwise).

Now resultant \(\vec{R} = (-P, -2P)\) acts at some point (x,y) on its line of action. Its moment about A is \(\vec{r} \times \vec{R} = x*(-2P) - y*(-P) = -2Px + Py\). This must equal total moment 5Ps (counterclockwise positive). So:

\(-2Px + Py = 5Ps \Rightarrow -2x + y = 5s\).

The line of action of the resultant is along the direction of \(\vec{R}\), so its equation is: \(\frac{x - x_0}{-P} = \frac{y - y_0}{-2P}\) or equivalently \(2x - y = \text{constant}\). From the moment equation, we have \(-2x + y = 5s\), so the line is \(y - 2x = 5s\).

We need intersection with AD produced. AD is the line x=0 from y=0 to y=s, and produced means beyond D (y > s) or beyond A? "AD produced" typically means extended beyond D, so x=0, y > s. Substitute x=0 into line equation: y = 5s. So intersection point E is (0, 5s).

Now AE is distance from A(0,0) to E(0,5s) = 5s. DE is distance from D(0,s) to E(0,5s) = 4s. So AE : DE = 5s : 4s = 5 : 4. Hence proved.

Solution (First part):

Let the tower height be h. The top of tower is at height h above ground. The bottom is at ground level. The body is at horizontal distance a from the tower. Let the body be at height y above ground (same for both bullets as they strike same point).

For bullet from bottom (elevation β): initial velocity u (same for both? Assume same speed). The equations of motion: horizontal distance = u cosβ * t = a, so time t = a/(u cosβ). Vertical distance: y = u sin