In the right \(\triangle\)CEB
\( \tan 30^{\circ} =\frac{C E}{B E} \)
\(\frac{1}{\sqrt{3}} =\frac{C E}{12} \)
\(C E =\frac{12}{\sqrt{3}} \)
\(=\frac{12 \sqrt{3}}{\sqrt{3} \sqrt{3}} \)
\(=\frac{12 \times 1.732}{3}=4 \times 1.732 \)
= 6.928
= 6.93 km
Vertical height between B and C = 6.93 km.

2x² - x - 1 = 0 here, a = 2, b = -1, c = -1
α + β = \(\frac {-b}{a} = \frac {-(-1)}{2} = \frac {1}{2}\), αβ = \(\frac {c}{a} = -\frac {1}{2}\)
2α + β, 2β + α
Sum of the roots 2α + β + 2β + α = 3(α + β) = \(3\left( \frac { 1 }{ 2 } \right) =\frac { 3 }{ 2 } \)
Product of the roots = (2α + β) (2β + α) = 4αβ + 2α² + 2β² + αβ
= 5αβ + 2(α² + β²) = 5αβ + 2[(α + β)² - 2αβ]
= \(5\left( -\frac { 1 }{ 2 } \right) +2\left[ \frac { 1 }{ 4 } -2\times -\frac { 1 }{ 2 } \right] \) = 0
The required equation is x² - (Sum of the roots)x + (Product of the roots) = 0
x² - \(\frac { 3 }{ 2 } x\) + 0 = 0 gives 2x² - 3x = 0

The coordinate of the point of contact P of the tangent line AB with the circle is point of intersection of line.
x + y − 5 = 0 and x − y = 0
solving, we get x = \(\frac { 5 }{ 2 } \) and y = \(\frac { 5 }{ 2 } \)
Therefore, the coordinate of the P\(\left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) \)

LHS = (AB)^T 
AB = \({ \left[ \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right] }_{ 2\times 3 }\times { \left[ \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right] }_{ 3\times 2 }\)
= \(\left[ \begin{matrix} 2-2+0 & -1+8+2 \\ 4+1+0 & -2-4+2 \end{matrix} \right] =\left[ \begin{matrix} 0 & 9 \\ 5 & -4 \end{matrix} \right] \)
(AB)^T = \({ \left[ \begin{matrix} 0 & 9 \\ 5 & -4 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 0 & 5 \\ 9 & -4 \end{matrix} \right] \) ....(1)
RHS = (B^TA^T)
B^T = \(\left[ \begin{matrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{matrix} \right] \), A^T = \(\left[ \begin{matrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{matrix} \right] \)
B^TA^T = \({ \left[ \begin{matrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{matrix} \right] }_{ 2\times 3 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{matrix} \right] }_{ 3\times 2 }\)
= \(\left[ \begin{matrix} 2-2+0 & 4+1+0 \\ -1+8+2 & -2-4+2 \end{matrix} \right] \)
B^TA^T = \(\left[ \begin{matrix} 0 & 5 \\ 9 & -4 \end{matrix} \right] \)...(2)}
From (1) and (2), (AB)^T = B^TA^T.
Hence proved.

The height of the tower AB = 50 m
Let the height of the tree CD = y and BD = x
From the diagram,\(\angle \)XAC = 30° = \(\angle \)ACM and\(\angle \) = XAD = 45° =\(\angle \) ADB
In right triangle ABD,
tan45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 50 }{ x } \) gives x = 50 m
In right triangle AMC,
tan30° = \(\frac { AM }{ CM } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { AM }{ 50 } \)[since DB = CM]
AM = \(\frac { 50 }{ \sqrt { 3 } } =\frac { 50\sqrt { 3 } }{ 3 } =\frac { 50\times 1.732 }{ 3 } \) = 28.87 m.
Therefore, height of the tree = CD = MB = AB − AM = 50 – 28.87 = 21.13 m

Let the assumed mean A = 34.5

Standard deviation \(\sigma=C \times \sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{N}-\left(\frac{\Sigma f_{i} d_{i}}{N}\right)^{2}} \)
\(\sigma=2 \times \sqrt{\frac{1052}{84}-\left(\frac{-158}{84}\right)^{2}} \)
\(\sigma=2 \times \sqrt{\frac{22092}{1764}-\frac{6241}{1764}} \)
\(\sigma=2 \times \sqrt{\frac{15851}{1764}}=2 \times \sqrt{8.98} \)
\(\sigma=2 \times 2.99 \simeq 5.99 \)
Standard deviation \(\sigma \simeq 5.99 \simeq 6\)

In the right \(\triangle\)ADB
\( \tan 20^{\circ} =\frac{A D}{D B} \)
\(0.3640 =\frac{A D}{8}\)
AD = 8 x 0.3640 = 2.91 km
Vertical height between A and B = 2.91 km

Let AB denote the height of the peak and be 'h'.
In ΔABC, 
tan 8^o = \(\frac { AB }{ BC } =\frac { h }{ m } \)
m = \(\frac { AB }{ BC } =\frac { h }{ m } \) ...(1)

In Δ ABC,
tan 40^o =\(\frac { AB }{ BC } =\frac { h }{ m } \)
m+1 = \(\frac { h }{ tan4 } \) ..(2)
From (1) and (2)
\(\frac { h }{ tan8 } +1=\frac { h }{ tan4 } \)
\(1=\frac { h }{ tan4 } -\frac { h }{ tan8 } \)
\(h\left[ \frac { tan8-tan4 }{ tan4tan8 } \right] =1\)
h = \(\frac { tan4\times tan8 }{ tan8-tan4 } \)
= 0.14 mile (approx)

Let CF be the height of the man; AD be the height of the window; BC is the distance between the observer and the house.
From the right triangle \(\triangle\)CBD
\( \tan 45^{\circ} =\frac{D B}{B C} \)
\(1 =\frac{D B}{5} \)
DB = 5m     ...(1)
From the right triangle CBA
\( \tan 60^{\circ} =\frac{A B}{C B} \)
\(\sqrt{3} =\frac{A D+D B}{5} \)
\(5 \sqrt{3} =\mathrm{AD}+5 \quad[\because \text { from }(1) D B=5 \mathrm{~m}] \)
\(\mathrm{AD} =5 \sqrt{3}-5=5(\sqrt{3}-1) \)
\(\mathrm{AD} =5(1.732-1) \)
\( {[\text {Given } \sqrt{3}=1.732] }\)
= 5 x 0.732 = 3.660
Height of the window = 3.66 m

Equation of line AB, A(0, 5) and B(4,1)
\(\frac { { y-y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { { x-x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac { y-5 }{ 1-5 } =\frac { x-0 }{ 4-0 } \)
4(y - 5) = -4x gives y - 5 = - x
x + y - 5 = 0

Area of tin sheet required
= C.S.A of cylinder + C.S.A of frustum
Cylinder:
Radius = 4 cm
Height = 10 cm
C.S.A  \(=2 \pi r h\ sq. units \)
\(=2 \times \frac{22}{7} \times 4 \times 10=\frac{1760}{7}\ sq. units\)
Frustum of a cone
r₁ = radius of top = 9 cm
r₂ = radius of bottom = 4 cm
Height = 22 - 10 = 12 cm
Slant height \(l =\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}} \)
\(=\sqrt{12^{2}+(9-4)^{2}} \)
\(=\sqrt{144+25}=\sqrt{169}=13 \mathrm{~cm} \)
C.S.A  \(=\pi\left(r_{1}+r_{2}\right) l \text { sq, units } \)
\(=\frac{22}{7}(9+4)(13) \)
\(=\frac{3718}{7} \text { sq. units } \)
Area of tin sheet \(=\frac{1760}{7}+\frac{3718}{7}\)
\(=\frac{5478}{7}=782.57 \mathrm{~cm}^{2}\)

Clearly In \(\triangle\)ABC, D, E, F are points on lines BC, CA, AB respectively using Ceva's theorem, we have
\(\frac{A E}{E C} \times \frac{C D}{D B} \times \frac{B F}{F A}=1\)       ...(1)
From the diagram it is clear that
AE = 3, EC = 4, CD = 10, DB = 3, FA = 5
Substituting these values in (1)
\( \frac{3}{4} \times \frac{10}{3} \times \frac{B F}{5} =1 \)
\(B F =\frac{1 \times 4 \times 3 \times 5}{3 \times 10}=2 \mathrm{~cm}\)

Since OP is the radius and PT tangent
\(\angle O P T=90^{\circ}\)
Applying Pythagoras theorem in \(\triangle\)OPT, we have
OT² = OP² + PT²
13² = 5² + PT²
PT² = 169 - 25
PT² = 144
PT = 12cm
Since the lengths of tangents drawn from an exterior point to a circle are equal.
AP = AE = x(say)
AT = PT - AP = (12-x)cm
Since AB is the tangent to the circle at E
\( \therefore \mathrm{OE} \perp \mathrm{AB} \)
\(\ \Rightarrow \angle O E A =90^{\circ} \)
\(\\ \Rightarrow \angle A E T =90^{\circ}\)
\( \\ \mathrm{AT}^{2} =\mathrm{AE}^{2}+\mathrm{ET}^{2}\)
[Applying Pythagoras theorem in \(\triangle\)AET ]
(12 - x)² = x² +(13 -5)²
144 - 24 + x² = x² +64
24x = 144 - 64
24x = 80
3x = 10
\( x=\frac{10}{3} \mathrm{~cm} \)
\(Similarly\ B E=\frac{10}{3} \mathrm{~cm}\)
AB = AE + BE
\( =\left(\frac{10}{3}+\frac{10}{3}\right) \mathrm{cm} \)
\(A B =\frac{20}{3} \mathrm{~cm}\)

Edge of cube = 7 cm
surface area of a cube = 6a² sq. units
= 6(7)²
= 294 cm²
radius of hemisphere = \(\frac{7}{2} \mathrm{~cm}\)
[Only C.S.A is considered as the hemisphere surmounted]
C.S.A of hemisphere \(=2 \pi r^{2} \text { sq. units } \)
\(=2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
= 77 Cm²
Surface area of = T.S.A of cube + C.S.A the solid of hemisphere area of circular region (bottom of hemisphere)

\(=294+77-\left(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right)\)
= 371 - 38.5
= 332.5 cm²

b=-42
a=49

\(\frac { sinA-sinB }{ cosA+cosB } +\frac { cosA-cosB }{ sinA+sinB } =0\)
\(\mathrm{LHS}=\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos D}{\sin A+\sin B} \)
\(=\frac{(\sin A-\sin B)(\sin A+\sin B)+(\cos A-\cos B)(\cos A+\cos B)}{(\cos A+\cos B)(\sin A+\sin B)} \)
\(=\frac{\left(\sin ^{2} A-\sin ^{2} B\right)+\left(\cos ^{2} A-\cos ^{2} B\right)}{(\cos A+\cos B)(\sin A+\sin B)} \)
\(=\frac{\sin ^{2} A-\sin ^{2} B+\cos ^{2} A-\cos ^{2} B}{(\cos A+\cos B)(\sin A+\sin B)} \)
\(=\frac{\left(\sin ^{2} A+\cos ^{2} A\right)-\left(\sin ^{2} B+\cos ^{2} B\right)}{(\cos A+\cos B)(\sin A+\sin B)} \)
\(=\frac{1-1}{(\cos A+\cos B)(\sin A+\sin B)} \)
\(=\frac{0}{(\cos A+\cos B)(\sin A+\sin B)}=0=\mathrm{RHS}\)

Let l, b and h₁ be the length, breadth and height of the cuboid. Also let us take r and h₂ be the radius and height of the cylinder.
Now, Volume of the box = Volume of the cuboid + \(\frac{1}{2}\) (Volume of cylinder)
\((l\times b\times { h }_{ 1 })+\frac { 1 }{ 2 } ({ \pi r }^{ 2 }{ h }_{ 2 })cu.units\)
\(=\left( 30\times 15\times 10 \right) +\frac { 1 }{ 2 } \left( \frac { 22 }{ 7 } \times \frac { 15 }{ 2 } \times 30 \right) \)
= 4500 + 2651.79 = 7151. 79
Therefore, Volume of the box = 7151.79 cm³

Let TR = y. Since, OT is perpendicular bisector of PQ
PR = QR = 4 cm
In\(\triangle\)ORP, OP² = OR² + PR²
OR² = OP² - PR²
OR² = 5² - 4² = 25 - 16 = 9 \(\Rightarrow\) OR = 3cm
OT = OR + RT = 3 + y          ..(1)
In \(\triangle\)PRT, TP² + TR² + PR²      ..(2)
and \(\triangle\)OPT we have, OT² = TP² + OP²
OT² = (TR² + PR²) + OP² (substitute for TP² from (2))
(3 + y)² = y² + 4² + 5² (substitute for OT from (1))
9 + 6y² + 16 + 25 therefore \(y=TR=\frac { 16 }{ 3 } \)
6y = 41 - 9 we get  \(y=\frac { 16 }{ 3 } \)
From (2), TP² = TR² + PR²
\(TP2=\left( \frac { 16 }{ 3 } \right) +4^{ 2 }=\frac { 256 }{ 9 } +16=\frac { 400 }{ 9 } \)so, \(TP=\frac { 20 }{ 3 } \)

Let r and R be the inner and outer radii of the hollow sphere.
Given that, inner diameter d = 14 cm; inner radius r = 7 cm; thickness = 1 mm = \(\frac{1}{10}\)cm
Outer radius R = 7 + \(\frac { 1 }{ 10 } =\frac { 71 }{ 10 } =7.1cm\)
Volume of hollow sphere \(=\frac { 4 }{ 3 } \pi \left( { R }^{ 3 }-{ r }^{ 3 } \right) cu.cm\)
\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } (357.91-343)=62.48cm^{ 3 }\)
But, weight of brass in 1 cm³ = 17.3 gm
Total weight = 17362 x 62.48 = 1080.90 gm
Therefore, total weight is 1080.90 grams.

Let A be the starting position and 'B' be the final position. C be the position south of A at 34 m distance.
Clearly C = 90^o in \(\triangle\)ACB
AC² + CB² = AB²
[By Pythagoras theorem]
34² + 41² = AB²
1156 + 1681 = AB²
2837 = AB²
AB = 53.26 m
Distance from A to B through the pond = 53.26 m
Distance through C = 34m + 41 m = 75 m
Difference = 75 - 53.26 = 21.74 m
21.74 m would be saved if it is possible to walk through the pond.

\(\frac { 4{ x }^{ 2 }y }{ 2{ x }^{ 2 } } \times \frac { 6x{ z }^{ 3 } }{ 20{ y }^{ 4 } } =\frac { { 3x }^{ 3 }z }{ 5{ y }^{ 3 } } \)

Let the internal and external radii of the hemispherical shell be r and R
respectively.
Given that, R = 5 m, r = 3 m
C.S.A. of the shell = 2\(\pi\)(R² + r²) sq. units
\(=2\times \frac { 22 }{ 7 } \times \left( 25+9 \right) =213.71\)
T.S.A. of the shell = \(\pi\)(3R² + r²) sq. units
\(=\frac { 22 }{ 7 } (75+9)=264\)
Therefore, C.S.A. = 213.71 m² and T.S.A. = 264 m².

In \(\triangle\) ABC we have DE || BC.
By Thales theorem, we have \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\frac { x }{ x-2 } =\frac { x+2 }{ x-1 } \) gives x(x - 1) = (x - 2)(x + 2)
When x = 4, AD = 4, DB = x - 2, AE + x + 2 = 6, EC = x - 1 = 3 
Hence, AB = AD + DB = 4 + 2 = 6, AC = AE + EC = 6 + 3 = 9
Therefore, AB = 6, AC = 9

Area of the patio = Area of the quadrilateral ABCD - Area of the swimming pool EFGH
Area of Quadrilateral ABCD
A(- 4,- 8), B (8, - 4), C (6, 10) and D ( - 10, 6)
\(\text { Area } =\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)-\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \)
\(= \frac{1}{2}[(-4-6)(-4-6)-(8+10)(-8-10)] \)
\(= \frac{1}{2}[100+324]=\frac{424}{2}=212 \text { sq. units } \)
Area of Quadrilateral EFGH
E (- 3, - 5), F (6,- 2),G (3, 7) and H (- 6, 4)
Area \(=\frac{1}{2}[(-3,-3)(-2-4)-(6+6)(-5-7)] \)
\(=\frac{1}{2}[36+144]=\frac{180}{2}=90 \text { sq. units } \)
Area of patio = Area of quadrilateral ABCD - Area of Quadrilateral EFGH
= 212 - 90 = 122 sq. units

Squaring both sides(\(\sqrt { y+1 } +\sqrt { 2y-5 } \) )² = 3²
y+1+2y-5+2(\(\sqrt { y+1 } +\sqrt { 2y-5 } \)) = 9
3y-4-9 = -2\(\sqrt { y+1 } +\sqrt { 2y-5 } \)
Again squaring both sides
(3y-13)²(-2\(\sqrt { y+1 } +\sqrt { 2y-5 } \))²
9y²-78y+169 = 4(y+1)(2y-5)
9y²-78y+169 = 4(2y²+2y-5y-5)
9y²-78y+169 = 8y²+8y-20y-20
9y²-78y+169-8y²+12y+20 = 0
y²-66y+189 = 0
y²-63-3y+189 = 0
y(y-63)-3(y-63) = 0
(y-63)(y-3) = 0
y = 63, 3