Speed of ship \(=36 \mathrm{~km} / \mathrm{h}\)
Time taken \(=2 \mathrm{~min}\)
Therefore Distance covered by ship in \(2 \mathrm{~min}=36 \times \frac{2}{60}=1.2 \mathrm{~km}\)
Distance of ship from foot of lighthouse =1.3-1.2
\(=0.1 \mathrm{~km} =100 \mathrm{~m}\)
Let angle of elevation of the ship to the top of the lighthouse \(=y^{\circ}\)
Then, \(\tan y=\frac{173.2}{100}=1.732\)
\(\Rightarrow \quad \tan y=\sqrt{3} {[\because 1.732=\sqrt{3}]} \)
\( \Rightarrow \quad y=60^{\circ} \)
\({\left[\because \tan \sqrt{3}=\tan 60^{\circ}\right]}\)
Now, \(y=x^{\circ} \quad[\because\) alternate interior angles are equal ]
Thus, \(x^{\circ}=60^{\circ}\)

Trigonometric ratio involving AB, BC, OD, OA and angles is tanθ. [Refer AB, BC, OA and OD from the figure.]
Distance travelled by the balloon OB = AB - OA
From the figure, OD = BC, and it can be calculated as
88.2 m - 1.2 m = 87 m --- (1)
In ΔAOD,
tan 60° = OD/OA
√3 = 87/OA
OA = 87 / √3
= 87 × √3 / √3 × √3
= (87 × √3) / 3
= 29√3 m

45°

Yes, 7 m and 10 m

Let AB is the building

∴ AB=60 m and CD is the light house.
ㄥEAC=30^o
and ㄥEAD=60^o
∴ ㄥADB=60^o
∴ AE||BD
In right ΔABD,
\(\frac { BD }{ AB } \)=cot 60⁰⇒ \(\frac { BD }{ 60 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ BD=\(\frac { 60 }{ \sqrt { 3 } } \)   m=20\(\sqrt { 3 } \) m.
∵ BD=AE
∴ AE=20\(\sqrt { 3 } \) m
Now in right ΔCEA, tan 300=\(\frac { CE }{ AE } \)
⇒ \(\frac { 1 }{ \sqrt { 3 } } =\frac { CE }{ 20\sqrt { 3 } } \)⇒ CE=20 m
(i) Difference between the heights of the light house and the building=CE=20 m
(ii) The distance between the light house and the building =BD=20\(\sqrt { 3 } \) m

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Consider the equation
\(\frac { 4 }{ x } -3=\frac { 5 }{ 2x+3 } ,x\neq 0,-\frac { 3 }{ 2 } \)
\(\Rightarrow \frac { 4-3x }{ x } =\frac { 5 }{ 2x+3 } \)
\(\Rightarrow (4-3x)(2x+3)=5x\)
\(\Rightarrow 8x+12-6x^{ 2 }-9x=5x\)
\(\Rightarrow 6x^{ 2 }+6x-12=0\)
\(\Rightarrow 6(x^{ 2 }+x-2)=0\)
\(\Rightarrow x^{ 2 }+x-2=0\)
\(\Rightarrow x^{ 2 }+2x-x-2=0\)
\(\Rightarrow x(x+2)-1(x+2)=0\)
\(\Rightarrow (x+2)(x-1)=0\)
\(\Rightarrow \) Either x+2=0 or x-1=0
\(\Rightarrow \) x=-2,1

If the equation  x²+kx+64=0 has real roots,then D \(\ge 0\)
\(\Rightarrow k^{ 2 }-4\times 64\ge 0\quad \Rightarrow k^{ 2 }\ge 256\Rightarrow k^{ 2 }\ge (16)^{ 2 }\)
\(\Rightarrow k\ge 16\)   \(\left[ \therefore k>0 \right] \) ...(i)
If the equation x^{2 - 8x+k=0  has real roots then D _{\(\ge \) 0} \(\Rightarrow 64-4k\ge 0\Rightarrow 4k\le 64\) _...(ii)}
From (i) and (ii) we get k=16

Let speed of the stream be x km/h
Speed of the boat in still water = I I km/h 
\(\therefore \) Upstream speed be (11-x) km/h and downstream speed be (11+x) km/h
Distance = 12 km
Time taken for downstream direction = \(\frac { 12 }{ 11+x } \) hours
Time taken for upstream direction = \(\frac { 12 }{ 11-x } \) hours
ATQ \(\frac { 12 }{ 11+x } +\frac { 12 }{ 11-x } =2\frac { 3 }{ 4 } \Rightarrow \frac { 12(11-x)+12(11+x) }{ (11+x)(11-x) } =\frac { 11 }{ 4 } \)
\(\Rightarrow \frac { 132+132 }{ 121-x^{ 2 } } =\frac { 11 }{ 4 } \Rightarrow 4\times 264=11(121-x^{ 2 })\)
\(\Rightarrow \frac { 4\times 264 }{ 11 } =121-x^{ 2 }\Rightarrow 4\times 24=121-x^{ 2 }\)
\(\Rightarrow x^{ 2 }=25\Rightarrow x=\pm 5\)
Hence speed of the stream is x=5 km/h

\(p^{ 2 }+x^{ 2 }+(p^{ 2 }-q^{ 2 })x-q^{ 2 }=0\)
Here a=p²,b=(p²-q²),x= c=q²
D= b²-4ac=(p²-q²)²-4Xp²-4Xp²X(-q²) = (6²+q²)² 
Now  x= \(x=\frac { -b\pm \sqrt { D } }{ 2a } ,\frac { -b-\sqrt { D } }{ 2a } \)
\(\Rightarrow x=\frac { -(p^{ 2 }-q^{ 2 })+\sqrt { (p^{ 2 }+q^{ 2 })^{ 2 } } }{ 2\times p^{ 2 } } ;x=\frac { (-p^{ 2 }-q^{ 2 })-\sqrt { (p^{ 2 }-q^{ 2 })^{ 2 } } }{ 2\times p^{ 2 } } \)
\(\Rightarrow x=\frac { q^{ 2 } }{ p^{ 2 } } ,-1\)

Let the numbers be x, x + 2
(x)²+(x+2)²=340
x²+x²+4+4x=340
\(\Rightarrow 2x^{ 2 }+4x-3369=0\)
On dividing by 2, we get
x²+2x-168=0
\(\Rightarrow (x+14)(x-12)=0\)
\(\Rightarrow x=12\)
The numbers are, 12, (12 + 2) i.e., 12, 14

Let AB=h m be height of the tree and AC is the part of the broken tree. As ㄥCDB=450 and BD=8 m
Consider rt . angled ΔCBD, we have
tan45^o=\(\frac { CB }{ DB } \)
1=\(\frac { CB }{ 8 } \)
⇒ CB=8 m
Also, cos450=\(\frac { DB }{ DC } \)
\(\frac { 1 }{ \sqrt { 2 } } =\frac { 8 }{ DC } \)
⇒ Dc=8\(\sqrt { 2 } \)
Total height of the tree AB=AC+CB
=CD+CB
=8\(\sqrt { 2 } \)m+8m
=8(\(\sqrt { 2 } \) +1)m

5 Km/h

3 cm, 9 cm

41 years, 11 years

Let AC = x m and CE = y m

In rt.ΔACB, tan60^o =\(\frac { BC }{ AC } \)
\(\sqrt { 3 } =\frac { 3600\sqrt { 3 } }{ x } \)
x = 3600 m
Now, In right ΔAED,
tan300 = \(\frac { DE }{ AE } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { 3600\sqrt { 3 } }{ 3600+y } \)
3600 + y = 10800
y = 7200 m
BD = CE
∴ BD = 7200 m
∴ Distance covered in 30 seconds
= 7200 m
So, Speed = \(\frac { 720 }{ 30 } \) = 240 m/s
=\(240\times \frac { 18 }{ 5 } \) = 864 km/hr

Let height of the cloud C from lake be h m . A is position of the point 60 m above the lake. D is the reflection of the cloud in lake .
Let AE=x m, CF=h m, CE=(h-60)m
DE=(60+h)m.
In right angled triangle AEC
\(\frac { AE }{ EC } \)=cot 30⁰
AE=(h-60)\(\sqrt { 3 } \)           .........(i)
In right angled triangle AED,
\(\frac { AE }{ ED } =cot60^{ 0 }\quad \Rightarrow \quad AE=\frac { h+60 }{ \sqrt { 3 } } \)            .............(ii)
From (i) and (ii), we get
\((h-60)\sqrt { 3 } =\frac { h+60 }{ \sqrt { 3 } } \)
⇒ 3h-180=h+60 ⇒ 2h=240 ⇒ h=120 m.
∴ Height of the cloud above the lake is 120 m.

Let ABC is the right-angled triangle.
Let BC=x units and AB=y units.
Also AC Hypotenuse=29 units

Now Perimeter=70 units
\(\Rightarrow\) x+y+29=70
\(\Rightarrow\) x+y=41
\(\Rightarrow\) y=41-x    ....(i)
Also, AC²=AB²+BC²
\(\Rightarrow\) (29)²=(y)²+x²
\(\Rightarrow\) 841=(41-x)²+x² [Using (i)]
\(\Rightarrow\) 841=1681+x²-82x+x²
\(\Rightarrow\) 2x²-82x+840=0
\(\Rightarrow\) x²-41x+420=0
\(\Rightarrow\) (x-20)(x-21)=0
\(\Rightarrow\)  x=20 or x=21
Wnen x=20, y=41-x=21 and when x=21, y=41-21=20
\(\therefore\)  Length of other sides are 20 units and 21 units.

\(2\left( \frac { 2x-1 }{ x+3 } \right) -3\left( \frac { x+3 }{ 2x-1 } \right) =5\)
Let \(\frac { 2x-1 }{ x+3 } =y\)
Given equation becomes \(2y-3\times \frac { 1 }{ y } =5\Rightarrow 2{ y }^{ 2 }-3=5y\Rightarrow 2{ y }^{ 2 }-5y-3=0\)
\(\Rightarrow 2{ y }^{ 2 }-6y+y+3=0\Rightarrow 2y(y-3)+1(y-3)=0\)
\(\Rightarrow \left( y-3 \right) (2y+1)=0\Rightarrow y=3,y=-\frac { 1 }{ 2 } \)
\(\Rightarrow \frac { 2x-1 }{ x+3 } =3\Rightarrow \frac { 2x-1 }{ x+3 } =-\frac { 1 }{ 2 } \Rightarrow 2x-1=3x+9or4x-2=-x-3\)
\(\Rightarrow \) -x = 10 or 5x = -1 \(\Rightarrow \) x = -10 or \(x=\frac { -1 }{ 5 } \)
\(\Rightarrow \) x = -10,\(\frac { -1 }{ 5 } \)