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The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.

Solution
                  
Let y be the height of the balloon from the second window
In \bigtriangleupAOB
\tan 30= \frac{y}{d} \; \; \left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )
d= y\sqrt{3}\; \cdots \left ( 1 \right )           \left ( \because \tan 30= 1\sqrt{3} \right )
In \bigtriangleupOCD
\tan 60= \frac{4+y}{d}
\sqrt{3}d= 4+y
d= \frac{4+y}{\sqrt{3}}\cdots \left ( 2 \right )
Equating equation (1) & (2) we get
y\sqrt{3}= \frac{4+y}{\sqrt{3}}
y\sqrt{3}\times \sqrt{3}= 4+y
3y-y= 4
2y= 4
y= \frac{4}{2}
y = 2
Height of balloon = 2 + 4 + y
= 2 + 4 + 2
= 8m

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A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h (1 + tan α cot β) metres.             

Solution.         According to the question :
                      
 

Let the height of the other house is X.
In \bigtriangleupDCF.
\tan \alpha = \frac{P}{B}= \frac{EC}{DC}
\tan \alpha =\frac{x-h}{DC}
DC= \frac{x-h}{\tan \alpha }\cdots \left ( 1 \right )
In \bigtriangleupDAB
. \tan \beta = \frac{P}{B}= \frac{DA}{AB}
\tan \beta = \frac{h}{AB}= \frac{h}{DC}         \left ( \because AB= DC \right )
\tan \beta = \frac{h}{DC}
DC= \frac{h}{\tan \beta }\cdots \left ( 2 \right )
from equation (1) and (2)
\frac{X-h}{\tan \alpha}= \frac{h}{\tan \beta}
X-h= \frac{\tan \alpha h}{\tan \beta }
X= \frac{\tan \alpha h+\tan \beta h}{\tan \beta}
X= \frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \beta}
separately divide
X= h\left ( \frac{\tan \alpha }{\tan \beta }+1 \right )
X= h\left ( 1+\tan \alpha \cot \beta \right )        \left ( \because \frac{1}{\tan \theta }= \cot \theta \right )
X= h\left ( 1+\tan \alpha \cot \beta \right )
Hence proved

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The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point, 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Solution.         According to question
                    

Let h is the height of the tower
In \bigtriangleupABE
\tan \theta = \frac{P}{B}
\tan 60^{\circ} = \frac{h}{AB}             \left ( \because \theta = 60^{\circ} \right )
\sqrt{3}= \frac{h}{AB}       \left ( \because \tan 60^{\circ}= \sqrt{3} \right )
AB= \frac{h}{\sqrt{3}}\: \cdots \left ( 1 \right )
In \bigtriangleupEDC
\tan \theta = \frac{P}{B}
\tan 45^{\circ} = \frac{h-10}{DC}
1= \frac{h-10}{AB}    \left ( \because AB-DC,\tan 45^{\circ}= 1 \right )
AB= h-10\: \cdots \left ( 2 \right )
from equation (1) and (2)
h-10= \frac{h}{\sqrt{3}}
\sqrt{3}h-\sqrt{3}10= h
\sqrt{3}h-h= \sqrt{3}\times 10
h\left ( \sqrt{3}-1 \right )= 10\sqrt{3}
h= \frac{10\sqrt{3}}{\sqrt{3}-1}
Hence the height of the tower
h= \frac{10\sqrt{3}}{\sqrt{3}-1}m

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A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show that\frac{p}{q}= \frac{\cos \beta -\cos \alpha }{\sin \alpha-\sin \beta }

Solution.         According to the question:-
                             

Here a and b be the angles of indication when the ladder at rest and when it pulled away from the wall
In \bigtriangleupAOB
\cos \alpha = \frac{OB}{AB}             cos \theta = \frac{Base}{Hypotenuse}
OB= AB\cos \alpha \cdots \left ( 1 \right )
\sin \alpha = \frac{AO}{AB}               \sin \theta = \frac{Perpendicular}{hypotenuse}

AO= AB\sin \alpha \cdots \left ( 2 \right )
Similarly In \bigtriangleupDOC
\cos \beta = \frac{OC}{DC}
OC= DC\cos \beta \cdots \left ( 3 \right )
\sin \beta = \frac{OD}{DC}
OD= DC\sin \beta \cdots \left ( 4 \right )
Now subtract equation (1) from (3) we get
OC – OB = DC cos\beta – AB cos\alpha
Here    OC – OB = P
and DC = AB because length of ladder remains \Rightarrow P = AB cos
\beta – AB cos\alpha
    P = AB (cos\beta – cos\alpha)         …(5)
Subtract equation (4) from (2) we get
AO – OD = AB sina – DC sin \beta
Here AO – OD = q
and AB = DC because length of ladder remains same
     \Rightarrow   q = AB sin \alpha – AB sin\beta
q = AB (sin \alpha – sin \beta)             …(6)
on dividing equation (5) and (6) we get
\frac{p}{q}= \frac{AB\left ( \cos \beta -\cos \alpha \right )}{AB\left ( \sin \alpha -\sin \beta \right )}
\frac{p}{q}= \frac{\cos \beta -\cos \alpha}{\sin \alpha -\sin \beta}
Hence proved

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From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects

 According to question

 

Let x and y are two objects and \beta and \alpha use the angles of depression of two objects.
In \bigtriangleupAOX
\tan \beta = \frac{h}{Ox}        \because \tan \theta = \frac{Perpendicular}{Base}
Ox= \frac{h}{\tan \beta }
Ox= h\cot \beta \cdots \left ( 1 \right )
In \bigtriangleupAOY
 \tan \alpha = \frac{h}{Oy}= \frac{h}{Ox+xy}\; \; \left ( \because Oy= Ox+xy \right )
Ox+xy= \frac{h}{\tan \alpha }
Ox+xy= h\cot \alpha
xy= h\cot \alpha- Ox
xy= h\cot \alpha- h\cot \beta  (from equation (1))
xy= h\left ( \cot \alpha- \cot \beta \right )
Hence the distance between two objects is

h\left ( \cot \alpha- \cot \beta \right ) .

 

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The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.

Solution.         According to the question
                        
Here 30 m is the length of tower AB.
Let h is the height of tower DC
Let the distance between them is x
In \bigtriangleupABC
\tan 60^{\circ}= \frac{30}{x}     \left [ \tan \theta = \frac{Perpendicular}{Base} \right ]
\sqrt{3}= \frac{30}{x}
x= \frac{30}{\sqrt{3}}\cdots \left ( 1 \right )
In \bigtriangleupBDC
\tan 30^{\circ}= \frac{h}{x}
\frac{1}{\sqrt{3}}= \frac{h}{30}\times \sqrt{3}    (using (1))
\frac{30}{\sqrt{3}}=h\sqrt{3}
\frac{30}{\sqrt{3}\times\sqrt{3} }= h
\frac{30}{3}= h
h = 10m
Hence the height of the second tower is 10
Distance between them

=\frac{30}{\sqrt{3}}m .

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Prove that

\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }= \frac{1-\sin \theta }{\cos \theta }
 

Solution  To prove :- \frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }= \frac{1-\sin \theta }{\cos \theta }

                    Taking left hand side

\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }

 =\frac{1+\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}{1+\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }}                 \begin{bmatrix} \because \because \sec \theta = \frac{1}{\cos \theta } & \\ \tan \theta = \frac{\sin \theta }{\cos \theta }& \end{bmatrix}
=\frac{\frac{\cos \theta+1-\sin \theta}{\cos \theta}}{\frac{\cos +1+\sin \theta}{\cos \theta}}
\frac{\cos \theta+1-\sin \theta}{\cos \theta+1+\sin \theta}
Multiply nominator and denominator by (1 – sin \theta)
=\frac{\left ( \cos \theta +1-\sin \theta \right )\left ( 1-\sin \theta \right )}{\left ( \cos \theta +1+\sin \theta \right )\left ( 1-\sin \theta \right )}
=\frac{\cos \theta-\cos \theta\sin \theta+1-\sin \theta-\sin \theta+\sin^{2} \theta}{\cos \theta-\sin \theta\cos \theta+1-\sin \theta+\sin \theta-\sin^{2} \theta}
=\frac{\cos ec\left ( 1-\sin \theta \right )+\left ( 1-\sin \theta \right )-\sin \left ( 1-\sin \theta \right )}{\cos \theta -\sin \theta \cos \theta+1-\sin ^{2} \theta}
=\frac{\left ( 1-\sin \theta \right )\left ( \cos \theta +1-\sin \theta \right )}{\cos \theta-\sin \theta\cos \theta+\cos^{2} \theta}       \left ( \because 1-\sin ^{2}\theta = \cos ^{2}\theta \right )
=\frac{\left ( 1-\sin \theta \right )\left ( \cos \theta +1-\sin \theta \right )}{\cos \theta\left ( \cos \theta+1-\sin \theta \right )}
= \frac{1-\sin \theta}{\cos \theta}
Hence proved

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If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ =\sqrt{a^{2}+b^{2}-c^{2}}

Solution.     Given:- asinθ + b cosθ = c
squaring both side we get
\left ( a\sin \theta +b\cos \theta \right )^{2}= c^{2}
a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta+2ab\sin \theta \cos \theta = c^{2}
\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )
\Rightarrow 2ab\sin \theta \cos \theta = c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \cdots \left ( 1 \right )
To prove : acosθ – b sinθ =\sqrt{a^{2}+b^{2}-c^{2}}
Taking left hand side : a cosθ – b sinθ and square it we get
\left ( a\cos \theta -b \sin \theta \right )^{2}
= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -2ab\cos \theta \sin \theta
\left [ \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right ]
= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -\left ( c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \right ) \text{[Using (1)]}
= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta - c^{2}+a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta
= a^{2}\left ( \cos ^{2}\theta+\sin ^{2}\theta \right )+b^{2}\left ( \sin ^{2}\theta+ \cos ^{2}\right )-c^{2}
= a^{2}+b^{2} - c^{2}      \left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )
Hence \left ( a\cos \theta -b\sin \theta \right )^{2}= a^{2}+b^{2}-c^{2}
\Rightarrow \left ( a\cos \theta -b\sin \theta \right )= \sqrt{a^{2}+b^{2}-c^{2}}
Hence proved.

 

 

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If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.

Solution.   Given :-sinθ + cosθ = p
and      secθ + cosecθ = q
To prove :-q (p2 – 1) = 2p
Taking left hand side
q.(p2– 1) =
Put value of q and p we get
\left ( \sec \theta +\cos ec\theta \right )\left [ \left ( \sin \theta +\cos \theta \right )^{2}-1 \right ]
\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )
= \left ( \frac{1}{\cos \theta +\frac{1}{\sin \theta }} \right )\left [ \left ( \sin^{2} \theta+\cos^{2} \theta+2 \sin \theta \cos \theta \right ) -1\right ]
\left ( \because \sec \theta = \frac{1}{\cos \theta } ,\cos ec\theta =\frac{1}{\sin \theta } \right )
= \frac{1}{\cos \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta -1 \right ]
+ \frac{1}{\sin \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cdot \cos \theta -1 \right ]
= \frac{\sin ^{2}}{\cos \theta }+\cos \theta+2\sin \theta -\frac{1}{\cos \theta}+\sin \theta+\frac{\cos^{2} \theta}{\sin \theta}+2\cos \theta-\frac{1}{\sin \theta}
= 3\cos \theta +3\sin \theta -\frac{1}{\cos \theta}+\frac{\left ( 1+\cos ^{2}\theta \right )}{\cos \theta }+\frac{\left ( 1+\sin ^{2}\theta \right )}{\sin \theta }-\frac{1}{\sin \theta }
\left ( \because \sin ^{2}\theta = 1-\cos ^{2}\theta \right )
\left ( \because \cos ^{2}\theta = 1-\sin ^{2}\theta \right )
= 3\cos \theta +3\sin \theta+\frac{1}{\cos \theta}\times \left ( -1+1-\cos^{2} \theta \right )+\frac{1}{\sin \theta }\times \left ( 1-\sin^{2} \theta-1 \right )
\left ( \because \sin^{2} \theta+\cos^{2} \theta= 1 \right )
= 3\cos \theta -\cos \theta +3\sin \theta -\sin \theta
= 2\cos \theta +2\sin \theta
= 2\left ( \cos \theta +\sin \theta \right )
2p (R.H.S)
Hence proved.

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If  tanθ + secθ =\l, then prove that

sec\theta=\frac{\l ^{2}+1}{2\l }

Solution.         Given : tanθ + secθ =l
There fore
\frac{\l ^{2}+1}{2\l }=\frac{\left ( \tan \theta +\sec \theta \right )^{2}+1}{2\left ( \tan \theta +\sec \theta \right )}
\frac{\tan^{2} \theta +\sec^{2} \theta+2\tan \theta \sec \theta+1}{2\left ( \tan \theta \right )+2\sec \theta }
\frac{\sec^{2} \theta +\sec^{2} \theta+2\tan \theta \sec \theta}{2\left ( \tan \theta+\sec \theta \right ) }       \left [ \because 1+\tan ^{2}\theta = \sec ^{2}\theta \right ]
\frac{2\sec^{2} \theta +2\tan \theta \sec \theta}{2\left ( \tan \theta+2\sec \theta \right ) }
\frac{2\sec \theta \left ( \sec \theta+\tan \theta \right ) }{2\left ( \tan \theta+\sec \theta \right ) }
= \sec \theta (R.H.S)
Hence proved

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