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Given, in \(\triangle ABC\), \(DE\parallel AC\)   [given]
So,  \(\frac { BE }{ EC } =\frac { BD }{ DA } \)  ... (i)
[ by basic proportionality theorem]
Also, \(DC\parallel AP\)    [given]
So, \(\frac { BC }{ CP } =\frac { BD }{ DA } \)  ... (ii)
[ by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac { BE }{ EC } =\frac { BC }{ CP } \)

Let aeroplane be at B and let the two ships at C and D, such that their angles of depression from B are \(30°\)and\(60°\), respectively.Then, angles of elevation of D and C from B are \(30°\)and \(60°\)respectively.

We have, AB=1200 m
Here, one side AB is common in both triangles.
Let AC=x m and CD= y m
In \(\Delta BAC\) , we have
\(tan\quad 60°=\frac { A }{ B } \Rightarrow \sqrt { 3 } =\frac { 1200 }{ x } \)
⇒ \(x=\frac { 1200 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 1200\sqrt { 3 } }{ 3 } =400\sqrt { 3 } \)  ....(i)
In \(\Delta BAD\) , We have
\(tan\quad 30°=\frac { AB }{ AD }=\frac { AB }{ DC+CA }\)    [∵ AD=DC+CA]
On putting the value of x from Eq.(i) in Eq. (ii) we get
\(y=1200\sqrt { 3 } -400\sqrt { 3 } \)
⇒ y = 1800\(\sqrt { 3 } \)
⇒ y = 800 x 1.732 \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow y=1385.6m\)
Hence, the distance between both ships is 1385. 6m.

(i) We have \(kx^{ 2 }+2x-3=0,x=2\)
Here,k is unknown since, x=2 is a solution of given equation,so it will satisfy the equation
On putting x=2 in given equation, we get 
\(k(2)^{ 2 }+2(2)-3=0\Rightarrow 4k+4-3=0\)
\(\Rightarrow 4k+1=0\)
\(\therefore k=\frac { -1 }{ 4 } \)
(ii) We have  \(x^{ 2 }+2ax-k=0,x=-a\) ,x=-a 
Here,k is unknown, Since , x=-a  is a solution of given equation , so it will satisfy the equation.
On putting x=-a in given equation , we get
\((-a)^{ 2 }+2a(-a)-k=0\Rightarrow a^{ 2 }-2a^{ 2 }-k=0\)
\(\Rightarrow -a^{ 3 }-k=0\Rightarrow k=-a^{ 2 }\)

\([7, -5 ]\)

\(\left [3\pm \sqrt {19}\over 5\right]\)

AD is the lighthouse, Find AD=?
In right ΔABD,
h=x
\(\frac { h }{ x } \)=tan45^o
\(\frac { h }{ 100-x } \)=tan30^o ......(ii)
Solve for h and x.
⇒ \(\frac { h }{ 100-x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } h\)=100-x
⇒ \(\sqrt { 3 } \)x=100-x   [Using eq (i)]
⇒ (\(\sqrt { 3 } \)+1)x=100 ⇒ x=\(\frac { 100 }{ \sqrt { 3 } +1 } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ (\sqrt { 3 } +1)(\sqrt { 3 } -1) } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ 2 } =50(\sqrt { 3 } -1)\)m
∴ h=height of lighthouse=\(50(\sqrt { 3 } -1)\)m

Let AP and DP be the position of the ladder whose length is 6 m.
In rt.ΔABP, \(\frac { BP }{ AP } =cos60^{ 0 }\Rightarrow \frac { BP }{ 6 } =\frac { 1 }{ 2 } \)
⇒ BP=1/2 x 6=3m
In rt, ΔDCP, \(\frac { PC }{ DP } =cos45^{ 0 }=\frac { PC }{ 6 } =\frac { 1 }{ \sqrt { 2 } } \)
⇒ PC=\(\frac { 6 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 6\sqrt { 2 } }{ 2 } =3\sqrt { 2 } m\)
Distance between two walls
=BP+PC=3+3\(\sqrt { 2 } \)
=3+3 x 1.41
=3(1+1.41)=7.23 m

\(\frac { x-1 }{ x-2 } +\frac { x-3 }{ x-4 } =\frac { 10 }{ 3 } \Rightarrow \frac { (x-1)(x-4)+(x-3)(x-2) }{ (x-2)(x-4) } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { x^{ 2 }-5x+4+x^{ 2 }-5x+6 }{ x^{ 2 }-6x+8 } =\frac { 10 }{ 3 } \Rightarrow 6x^{ 2 }-30x+30=10x^{ 2 }-60x+80\)
\(\Rightarrow 4x^{ 2 }-30x+50=0\Rightarrow 2x^{ 2 }-15x+25=0\)
\(\Rightarrow 2x^{ 2 }-10x-5x+25=0\Rightarrow 2x(x-5)-5(x-5)=0\)
\(\Rightarrow (x-5)(2x-5)=0\Rightarrow x=5\) or \(\frac { 5 }{ 2 } \)

Let the original speed of train be x km/hr. Then,
Increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed to cover 360 km = \(\frac{360}{x} \mathrm{hr}\)
Time taken by the train under increased speed to cover 360 km = \(\frac{360}{x+5} \mathrm{hr}\)
Therefore,
\( \frac{360}{x}-\frac{360}{x+5}=1 \)
\(\frac{360(x+5)-360 x}{x(x+5)}=1 \)
\(\frac{360 x+1800-360 x}{x^{2}+5 x}=1 \)
\(\frac{1800}{x^{2}+5 x}=1\)
1800 = x² + 5x
x² + 5x - 1800 = 0
x² - 40x + 45x - 1800 = 0
x(x - 40) + 45(x - 40) = 0
(x - 40)(x + 45) = 0
x - 40 = 0
x = 40
Or
x + 45 = 0
x = -45
But, the speed of the train can never be negative.
Hence, the original speed of train is x = 40 km/hr.