Let CD be the hill and the man is in A.
\(\angle E A D=60^{\circ} ; \angle B C A=30^{\circ}\)
\(\text { In } \triangle A E D\)
\(\tan 60^{\circ} =\frac{D E}{E A} \)
\(\sqrt{3} =\frac{h}{x} \)
\(h =\sqrt{3} x \)
\(\text { In } \triangle A B C\)
\(\tan 30^{\circ} =\frac{A B}{B C} \)
\(\frac{1}{\sqrt{3}} =\frac{10}{x} \)
\(x =10 \sqrt{3} \)
\(\text { From (1) and (2) } \quad \mathrm{h}=\sqrt{3}(10 \sqrt{3})\)
\([\because x=10 \sqrt{3} \text { in (1) }]\)
\(=10 \times 3=30 \)
\(\mathrm{CD} =\mathrm{CE}+\mathrm{ED} \)
\(=10+30 \)
\(=40 \mathrm{~m}\)
Distance of the hill from the ship is \(10 \sqrt{3} m\)
Height of the hill = 40 m

In the figure A is the stationary helicopter F is its reflection in the lake.
In right \(\Delta\)AED
\(\tan 30^{\circ} =\frac{A E}{D E} \)
\(\tan 30^{\circ} =\frac{1}{\sqrt{3}}=\frac{A E}{D E} \)
\(\frac{1}{\sqrt{3}} =\frac{x-100}{y} \)
\(y =(x-100) \sqrt{3}\)
\(\text { In right } \triangle D E F\)
\(\tan 60^{\circ} =\frac{E F}{D E} \)
\(\frac{x+100}{y} =\sqrt{3} \)
\(\sqrt{3} y =x+100 \)
\(y =\frac{(x+100)}{\sqrt{3}} \)
\(\text { From (1) and (2) we have }\)
\(\frac{x+100}{\sqrt{3}}=\sqrt{3}(x-100)\)
\(\sqrt{3} \times \sqrt{3}(x-100) =x+100 \)
\(3(x-100) =x+100 \)
\(3 x-300-x-100 =0 \)
2x = 400
x = 200
therefore height of the helicopter = 200 m.

In the figure, let BC be the tower and CD be the pole.
Let BC = 'x' m and AB = 'y' m
\(\text { In right } \triangle A B C\)
\(\frac{B C}{A B} =\tan 45^{\circ}=1 \)
\(B C =A B \)
\(y =x\)
\(\text { In right } \triangle A B D\)
\(\frac{B D}{A B}=\tan 60^{\circ}=\sqrt{3} \)
\(\frac{x+5}{y}=\sqrt{3} \)
\(y \sqrt{3}=x+5 \)
\(x \sqrt{3} =x+5 \quad[\because x=y \text { from (1) }\)
\(\sqrt{3} x-x =5 \)
\((\sqrt{3}-1) x =5 \)
\(x =\frac{5}{\sqrt{3}-1} \)
\(=\frac{5}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \)
\(=\frac{5(\sqrt{3}+1)}{3-1}=\frac{5(1.732+1)}{2} \)
\(=\frac{5}{2} \times 2.732=6.83 \mathrm{~m} \)
\(\therefore \text { Height of the tower is } 6.83 \mathrm{~m} \text {. }\)

Let CE be the building and AB be the lamp post
CE = 60m
\(\text { In right } \triangle B C E\)
\(\frac{B D}{A B}=\tan 60^{\circ}=\sqrt{3} \)
\(\frac{x+5}{y}=\sqrt{3} \)
\(y \sqrt{3}=x+5 \)
\(x \sqrt{3}=x+5 \quad[\because x=y \text { from }(1)] \)
\(\text { In right triangle } \triangle A D E\)
\(\tan 30^{\circ} =\frac{D E}{A D} \)
\(\frac{1}{\sqrt{3}} =\frac{D E}{20 \sqrt{3}} \)
\([\text { From }(1) \text { and } \mathrm{BC}=\mathrm{DE}]\)
\(\mathrm{DE}=\frac{20 \sqrt{3}}{\sqrt{3}}=20 \mathrm{~m}\)
Height of the lamp post = AB - CD
CE-DE
= 60m - 20m = 40m
Distance between the lamp post and building
\(=20 \sqrt{3} m \)
\(=20 \times 1.732 \mathrm{~m} \)
\(=34.64 \mathrm{~m} \)

Let A represents the position of the observer
AB = 100
\(\tan 60^{\circ} =\frac{A B}{B C} \)
\(\sqrt{3} =\frac{100}{B C} \)
\(B C =\frac{100}{\sqrt{3}} \)
\(=\frac{100 \sqrt{3}}{3} \)
\(=\frac{100 \times 1.732}{3}=57.73 \mathrm{~m} \)
\(\text { In right triangle } \triangle A B D\)
\(\frac{A B}{B D} =\tan 30^{\circ}=\frac{1}{\sqrt{3}} \)
\(\frac{1}{\sqrt{3}} =\frac{100}{B D} \)
\(B D =\sqrt{3} \times 100 \)
\(=1.732 \times 100=173.2 \)

Let AB is the tower AC and AD are shadows when the angle of elevation of the sun are 45^o and 30^o respectively
CD = 10

\(\text { In } \triangle C A B\)
\(\tan 45^{\circ} =\frac{A B}{A C} \)
\(1 =\frac{A B}{A C} \)
\(A C =A B \)
\(\text { In the right triangle } \Delta D A B\)
\(\tan 30^{\circ} =\frac{A B}{A D} \)
\(\frac{1}{\sqrt{3}} =\frac{A B}{A C+C D}=\frac{A B}{A C+10} \)
\(A C+10 =A B \sqrt{3} \)
\(\text { Using (1) } \mathrm{AC}+10=\sqrt{3} \mathrm{AC}\)
\(\sqrt{3} A C-A C =10 \)
\((\sqrt{3}-1) A C =10 \)
\(A C =\frac{10}{\sqrt{3}-1} \)
\(A C =\frac{10}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \)
\(=\frac{10(\sqrt{3}+1)}{3-1} \)
\(=\frac{10(\sqrt{3}+1)}{2}=5(\sqrt{3}+1)\)
\(=5(1.732+1)=13.65 \mathrm{~m}\)
\(\therefore \text { Height of the tower }=13.65 \mathrm{~m} \text {. }\)

Let AC be the tree.
BD be the broken part of the tree.
BD = BC
\(\text { In the right triangle } \triangle A B D\)
\(\tan 30^{\circ} =\frac{A B}{A D} \)
\(\frac{1}{\sqrt{3}} =\frac{A B}{30} \)
\(A B =\frac{30}{\sqrt{3}} m \)
\(\text { Also In } \triangle A B D\)
\(\cos 30^{\circ} =\frac{A D}{B D} \)
\(\frac{\sqrt{3}}{2} =\frac{30}{B D} \)
\(B D =\frac{2 \times 30}{\sqrt{3}} m=\frac{60}{\sqrt{3}} \mathrm{~m} \)
\(\therefore \text { Height of the tree }=\mathrm{AB}+\mathrm{BC}\)
\(=A B+B D \)
\(=\frac{30}{\sqrt{3}}+\frac{60}{\sqrt{3}} \)
\(=\frac{30+60}{\sqrt{3}}=\frac{90}{\sqrt{3}} \)
\(=\frac{30 \times \sqrt{3} \times \sqrt{3}}{\sqrt{3}}=30 \sqrt{3} \mathrm{~m} \)

\(\text { Given } \ \tan A=n \tan B\)
\(\Rightarrow \tan B=\frac{1}{n} \tan A \)
\(\Rightarrow \frac{1}{\tan B}=\frac{n}{\tan A} \)
\(\Rightarrow \cot B=\frac{n}{\tan A}\)
\(\text { Also } \sin \mathrm{A}=\mathrm{m} \sin \mathrm{B}\)
\(\Rightarrow \sin B=\frac{1}{m} \sin A \)
\(\Rightarrow \frac{1}{\sin B}=\frac{m}{\sin A} \)
\(\Rightarrow \operatorname{cosec} B=\frac{m}{\sin A} \)
\(\text { We know that } \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\)
\(\text { Now } \operatorname{cosec}^{2} \mathrm{~B}-\cot ^{2} \mathrm{~B}=1\)
\(\Rightarrow \frac{m^{2}}{\sin ^{2} A}-\frac{n^{2}}{\tan ^{2} A}=1 \)
\(\Rightarrow \frac{m^{2}}{\sin ^{2} A}-n^{2} \frac{\cos ^{2} A}{\sin ^{2} A}=1 \)
\(\frac{m^{2}-n^{2} \cos ^{2} A}{\sin ^{2} A}=1 \)
\(m^{2}-n^{2} \cos ^{2} A =\sin ^{2} A \)
\(m^{2} =\sin ^{2} A+n^{2} \cos ^{2} A \)
\(m^{2} =1-\cos ^{2} A+n^{2} \cos ^{2} A \)
\(\Rightarrow m^{2}-1 =n^{2} \cos ^{2} A-\cos ^{2} A \)
\(\Rightarrow m^{2}-1 =\left(n^{2}-1\right) \cos ^{2} A \)
\(\Rightarrow \frac{m^{2}-1}{n^{2}-1} =\cos ^{2} A \)
\(\cos ^{2} A=\frac{m^{2}-1}{n^{2}-1}\)

\(\text { Given } \ \operatorname{cosec} \theta-\sin \theta=m\)
\(\Rightarrow \frac{1}{\sin \theta}-\sin \theta =m \)
\(\frac{1-\sin ^{2} \theta}{\sin \theta} =m \)
\(\frac{\cos ^{2} \theta}{\sin \theta} =m \)
\(\text { Also } \sec \theta-\cos \theta=n\)
\(\frac{1}{\cos \theta}-\cos \theta=n \)
\(\Rightarrow \frac{1-\cos ^{2} \theta}{\cos \theta}=n \)
\(\Rightarrow \frac{\sin ^{2} \theta}{\cos \theta}=n \)
\(\mathrm{LHS}=\left(\mathrm{m}^{2} \mathrm{n}\right)^{2 / 3}+\left(\mathrm{mn}^{2}\right)^{2 / 3}\)
\(=\left(\cos ^{3} \theta\right)^{\frac{2}{3}}+\left(\sin ^{3} \theta\right)^{\frac{2}{3}} \)
\(=\cos ^{2} \theta+\sin ^{2} \theta \)
\(=1=\mathrm{RHS} \)

\(\text { Given } \ a \cos \theta+b \sin \theta=m\)
\(a \sin \theta-b \cos \theta=n\)
\(\mathrm{RHS}=\mathrm{m}^{2}+\mathrm{n}^{2}\)
\(=(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}\)
\(=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a b \sin \theta \cos \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-a b \sin \theta \cos \theta\)
\(=a^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\)
\(=\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{LH} \mathrm{S}\)

\(\mathrm{LHS}=\sec \theta+\tan ^{3} \theta \operatorname{cosec} \theta\)
\(=\sec \theta\left\{\frac{\sec \theta+\tan ^{3} \theta \operatorname{cosec} \theta}{\sec \theta}\right\}\)
\([\because \text { Multiplying and dividing by } \sec \theta]\)
\(=\sec \theta\left\{\frac{\frac{1}{\cos \theta}+\tan ^{3} \theta \operatorname{cosec} \theta}{\frac{1}{\cos \theta}}\right\}\)
\(=\sec \theta\left\{\frac{\frac{1+\tan ^{3} \theta \cos \theta \cdot \operatorname{cosec} \theta}{\cos \theta}}{\frac{1}{\cos \theta}}\right\}\)
\(=\sec \theta \frac{\left(1+\tan ^{3} \theta \operatorname{cosec} \theta \sec \theta\right.}{\cos \theta} \times \frac{\cos \theta}{1}\)
\(=\sec \theta\left[1+\tan ^{3} \theta \frac{\cos \theta}{\sin \theta}\right] \)
\(=\sec \theta\left(1+\tan ^{3} \theta \cot \theta\right) \)
\(=\sqrt{1+\tan ^{2} \theta}\left\{1+\tan ^{3} \theta \times \frac{1}{\tan \theta}\right\} . \)
\(=\sqrt{1+\tan ^{2} \theta}\left(1+\tan ^{2} \theta\right) \)
\(=\left(1+\tan ^{2} \theta\right)^{\frac{1}{2}}\left(1+\tan ^{2} \theta\right) \)
\(=\left(1+\tan ^{2} \theta\right)^{\frac{3}{2}}=\left(1+\left(1-a^{2}\right)\right)^{\frac{3}{2}} \)
\(=\left(1+1-a^{2}\right)^{\frac{3}{2}}=\left(2-a^{2}\right)^{\frac{3}{2}}=\mathrm{RHS} \)