109.6 m

\(6\sqrt { 3 } m\)

Let p is the position of the kite with height 45 m from the point Q on the ground. Let R is the other point on the ground to which kite is temporarily tied
AS ㄥPRQ=60^o and PQ=45 m (given)
Consider art. ΔPQR, we have
sin 60^o =\(\frac { PQ }{ PR } \)
\(\frac { \sqrt { 3 } }{ 2 } =\frac { 45 }{ PR } \)
PR=\(\frac { 90 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)
PR=\(\frac { 90\sqrt { 3 } }{ 3 } \)
=30\(\sqrt { 3 } \)
Hence, the length of the string is 30m.\(\sqrt { 3 } \) m.

The angle of depression and the angle of elevation from an object on the ground to an object in the air equal.

In right \(\Delta ABC\)
\(tan\theta =\frac { AB }{ BC } \) 
\(\Rightarrow tan\theta =1 \left( \because AB=BC \right) \) 
\(\Rightarrow \theta ={ 45 }^{ o }\) 

Hint Let a man is standing on the deck of a ship at point A such that \(A B=10 \mathrm{~m}\) and let CD be the hill.
Then, \(\angle E A D=60^{\circ}\)
and \(\angle C A E=\angle B C A=30^{\circ}\) [alternate angles]
Let \(B C=x \mathrm{~m}=A E\) and \(D E=h \mathrm{~m}\)
In right angled \(\triangle A E D\),
\(\tan 60^{\circ}=\frac{P}{B}=\frac{D E}{E A}=\frac{h}{x}\)
In right angled \(\triangle A B C\),
\(\tan 30^{\circ}  =\frac{A B}{B C}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}  =\frac{10}{x}  {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]}\)
Ans. Distance of the hill from the ship is \(10 \sqrt{3} \mathrm{~m}\) and height of the hill is 40 m .

Let AB = h km be the height of the mountain. Let C be a point at a distance of x km from the base of the mountain such that \(\angle\)ACB = 30\(°\) and let D be a point at a distance of 10 km from C along the same line. Then \(\angle\)ADB = 15\(°\) and AD = AC + DC = (x + 10) km

\(In\quad \Delta BAC,\quad tan\quad 30°=\frac { P }{ B } =\frac { AB }{ AC } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x } \left[ \because tan 30°=\frac { 1 }{ \sqrt { 3 } } \right] \)
\(\Rightarrow x= h\sqrt { 3 } \)  ......(i)
\(In \Delta BAD, tan 15°=\frac { AB }{ AD } \)
\(\Rightarrow 0.27=\frac { h }{ x+10 } \left[ given, tan 15°=0.27 \right] \)
\(\Rightarrow\)  0.27(x + 10) = h  ......(ii)
On putting x = \(\sqrt { 3 } h\) from Eq. (i) in Eq. (ii), we get
\(0.27(\sqrt { 3 } h+10)=h\)
\(\Rightarrow 0.27\times \sqrt { 3 } h+0.27\times 10=h\)
\(\Rightarrow h(1-0.27\times \sqrt { 3 } )=0.27\times 10\)
\(\Rightarrow h(1-0.27\times 1.732)=2.7\)   \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow\) h(1-0.47) = 2.7
\(\Rightarrow 0.53h=2.7\Rightarrow h=\frac { 2.7 }{ 0.53 } =5.09\approx 5km\)
Hence, the height of mountain is 5 km.

34.64 m; 20 m

Let P be the position of the aeroplane. Then, PM=200m and let A and B be two points on the two banks of a river such that the angles of depression at A and B are \({ 60 }^{ \circ }\)and \({ 45 }^{ \circ }\) , respectively. 
Let  Am=x m and BM= y m.
Then, \(\angle XPB=\angle MBP={ 45 }^{ \circ }\)  [altenatives angles]
and  \(\angle XPB=\angle MBP={ 60 }^{ \circ }\)  [altenatives angles]
In right angled  \(\Delta AMP,\) \(\tan { { 60 }^{ \circ } } =\frac { PM }{ AM } \)
\(\Rightarrow \sqrt { 3 } =\frac { 200 }{ x } \Rightarrow 200=\sqrt { 3x } \)
\(\Rightarrow x=\frac { 200 }{ \sqrt { 3 } } m \)
In right angled  \(\Delta BMP,\)
\(\tan { { 45 }^{ \circ } } =\frac { PM }{ BM } \Rightarrow 1=\frac { 200 }{ y }\)
\(\Rightarrow y=200m \)
Now, width of the river, AB=BM+MA
\(\Rightarrow AB=x+y=\frac { 200 }{ \sqrt { 3 } } +200\)
\( =200\left( \frac { 1 }{ \sqrt { 3 } } +1 \right)\)
\(=200(1.5773)=315.46m. [\because \sqrt { 3 } =1.732]\)
Hence, the width of the river is 315.46m.

Let AB be the height of the balloon represented by h in above a straight road, P and Q be the positions of two cars 100 m apart, such that ㄥAPB=45^o, ㄥAQB=60^o, PQ=100 m.
Consider rt. ㄥAQB=60^o, PQ=10^o m.
\(\frac { AB }{ QB } \)=tan 60^o
AB=\(\sqrt { 3 } \) QB
QB=\(\frac { AB }{ \sqrt { 3 } } =\frac { h }{ \sqrt { 3 } } \) .......(i)
Consider rt.ㄥed ΔPBA we have
\(\frac { AB }{ PB } \)=tan 45^o
AB=PB
AB=PQ+QB
AB=100+QB
h=100+\(\frac { h }{ \sqrt { 3 } } \) [using (i)]
h\(\left( 1-\frac { 1 }{ \sqrt { 3 } } \right) \)=100
h=\(\frac { 100\sqrt { 3 } }{ \sqrt { 3 } -1 } =\frac { 100\sqrt { 3 } (\sqrt { 3 } +1) }{ (\sqrt { 3 } -1)(\sqrt { 3 } +1) } \)
=\(\frac { 100(3+\sqrt { 3 } ) }{ 2 } \)
=\(50(3+\sqrt { 3 } )\)m

Let a man is standing on the deck of a ship at point A such that AB = 10 m and let CD be the hill.
Then, ∠EAD = 60° and ∠CAE = ∠BCA = 30° [alternate angles]
Let BC = x m = AE and DE = h m

In right angled ΔAED, 
\(\tan 60^{\circ}=\frac{P}{B}=\frac{D E}{E A}=\frac{h}{x}\)
In right angled ΔABC,
\(\tan 30^{\circ}=\frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{x}\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]\)
distance of the hill from the ship is 10\(\sqrt3\) m and height of the hill is 40 m.

Here, AB is the chimney, CD the observer and ∠ ADE the angle of elevation (see Figure). In this case, ADE is a triangle, right-angled at E and we are required to find the height of the chimney.
We have AB = AE + BE = AE + 1.5 and DE = CB = 28.5 m
To determine AE, we choose a trigonometric ratio, which involves both AE and DE. Let us choose the tangent of the angle of elevation.

Now, \(\tan 45^{\circ}=\frac{\mathrm{AE}}{\mathrm{DE}}\) 
i.e., \(1=\frac{\mathrm{AE}}{28.5}\) 
Therefore, AE = 28.5
So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m.

Let SQ be the height and PQ be the shadow of a man.
According to the question, SQ = PQ

Again, let the angle of elevation of the sun be \(\theta \) 
In right angled \(\\ \Delta PQS\)
\(tan\theta =\frac { perpendicular }{ Base } =\frac { QS }{ PQ }\)
\( \Rightarrow tan\theta =\frac { QS }{ QS } \left[ \because PQ = QS \right]\)
\( \\ \Rightarrow tan\theta =1=tan 45° \left[ \because tan 45°=1 \right] \)
\( \therefore \theta =45°\)
Hence, the angle of elevation of the sun is 45\(°\)

Let BC = 50 m be the height of the tower and AD=h m be the height of the building. Angle of elevation of the top of a building from the foot of tower is \(\angle DBA=30°\) and angle of elevation of the top of tower from the foot of building \(\angle CAB=60°\)
Also, let AB = x m be the distance between foot of the tower and building.
 In right angled \(\Delta BAD\),
\(tan 30°=\frac { AD }{ AB } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x }\)
\( \Rightarrow h=\frac { x }{ \sqrt { 3 } } \)
Again, in right angled \(\Delta CBA\),
\(tan 60°=\frac { BC }{ AB } \)
\(\Rightarrow \sqrt { 3 } =\frac { 50 }{ x } \Rightarrow x=\frac { 50 }{ \sqrt { 3 } } m\)
On putting \(x=\frac { 50 }{ \sqrt { 3 } } \) in Eq. (i), we get
\(h=\frac { 50 }{ \sqrt { 3 } } \times \frac { 1 }{ \sqrt { 3 } } =\frac { 50 }{ 3 } =16\frac { 2 }{ 3 } m\)
Hence, the height of the building is \(16\frac { 2 }{ 3 } m\)
 

16 minutes 24 seconds

(i) (d): Clearly, \(\angle\)DAC = 60°
So,in \(\Delta\)ADC, we have
\(\tan 60^{\circ}=\frac{C D}{A D} \Rightarrow \sqrt{3}=\frac{6}{A D} \)
\(\Rightarrow A D=\frac{6}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): Clearly, \(\angle\)DBC = 30°
So, in \(\Delta\)BDC,we have
\(\tan 30^{\circ}=\frac{C D}{B D} \)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{6}{B D} \)
\(\Rightarrow B D=6 \sqrt{3} \mathrm{~m}\)
(iii) (b): Width of the river = AB = AD + BD
\(\begin{array}{l} =\frac{6}{\sqrt{3}}+6 \sqrt{3} \\ =6\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)=6\left(\frac{4}{\sqrt{3}}\right)=\frac{24}{\sqrt{3}} \mathrm{~m}=13.87 \mathrm{~m} \end{array}\)
(iv) (a): The angle of elevation and angle of depression are always acute angles.
(v) (c): In \(\Delta\)BCD,if BD = 21m, then
\( \tan 30^{\circ}=\frac{C D}{B D} \)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{C D}{21} \Rightarrow C D=\frac{21 \sqrt{3}}{3}=7 \sqrt{3} \mathrm{~m}\)

(i) (c)
(ii) (a): Let AB be the height of pedestal.
\(\text { In } \Delta A B C, \)
\(\tan 30^{\circ}=\frac{A B}{B C} \)
\(\Rightarrow \quad A B=\frac{25}{\sqrt{3}}=\frac{25}{1.73}=14.45 \mathrm{~m}\)

(iii) (d): Let x be the distance between Arun and Pankaj.

\(\text { In } \Delta A B \dot{D}, \tan 45^{\circ}=\frac{A B}{B D}\)
\(\Rightarrow B D=30 \mathrm{~m}\)
\(\text { Now, in } \Delta \overline{A B C} \text { , }\)
\(\tan 30^{\circ}=\frac{A B}{B C}\)
\(\Rightarrow \frac{30}{30+x}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow x=30(\sqrt{3}-1)=30 \times 0.73=21.9 \mathrm{~m}\)
(iv) (c): \(\text { In } \triangle A R S\)

\(\sin 30^{\circ}=\frac{R S}{A S} \)
\(\Rightarrow \quad \frac{12}{A S} =\frac{1}{2} \Rightarrow A S=12 \times 2=24 \mathrm{~m}\)
(v) (c): \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 60^{\circ}\)

\(\begin{array}{l} \Rightarrow \frac{A B}{15}=\sqrt{3} \\ \Rightarrow A B=15 \times 1.73=25.95 \mathrm{~m} \end{array}\)

(i) (c) : Since AE || FD
\(\therefore\) \(\angle\)EAD = \(\angle\)ADF = 30° [Alternate interior angles]

(ii) (b): Since, AE || BC
\(\therefore\) \(\angle\)EAC = \(\angle\)ACB = 60° [Alternate interior angles]
(iii) (a) : In \(\Delta\)ABC,
\(\begin{array}{l} \tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{50}{B C} \\ \Rightarrow \quad B C=\frac{50}{\sqrt{3}}=28.90 \mathrm{~m} \end{array}\)
(iv) (c): In \(\Delta\)ADF, \(\tan 30^{\circ}=\frac{A F}{F D}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{A B-B F}{F D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{50-C D}{\frac{50}{\sqrt{3}}}\)
\(\left[\because F D=B C=\frac{50}{\sqrt{3}}\right] \)
\(\Rightarrow \frac{50}{3}=50-C D \Rightarrow C D=50-\frac{50}{3}=\frac{100}{3}=33.33 \mathrm{~m}\)
(v) (d)