In the right triangle GOH, cos 24° = \(\frac { OG }{ GH } \)
0.9135 = \(\frac { OG }{ 250 } \); OG = 228.38 km
Distance of H to the north of G = 228.38 km.

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Let AB be the height of the TV tower.
CD = 20 m.
Let BC be the width of the canal.
In the right angled ΔABC, tan58^o  \(=\frac{A B}{B C}\)
\(1.6003=\frac{A B}{B C}\)       ...(1)
In the right angled ΔABD, tan30° = \(\frac{A B}{B D}=\frac{A B}{B C+C D}\)
\(\frac{1}{\sqrt{3}}=\frac{A B}{B C+20}\)      ...(2)
Dividing (1) by (2) we get \( \frac{1.6003}{\frac{1}{\sqrt{3}}} =\frac{B C+20}{B C} \)
\(B C =\frac{20}{1.7717}=11.29 \mathrm{~m} \)    ...(3)
\(1.6003=\frac{A B}{11.29}\)   [from (1) and (3)]
AB = 18.07
Hence, the height of the tower is 18.07 m and the width of the canal is 11.29 m.

Let AC be the height of the tower.
Let AB be the height of the building.
Then, AC = h metres, AB = 30m
In right triangle CBP,\(\angle \)CPB = \(60°\)
tan\( \theta \) = \(\frac { BC }{ BP } \)
tan \(60°\) \(\frac { AB+AC }{ BP } so,\sqrt { 3 } =\frac { 30+h }{ BP } \)      ...(1)
In right triangle ABP,\(\angle \)APB = \(45°\) 
\(tan\theta =\frac { AB }{ BP } \) 
tan\(45°\)=\(\frac { 30 }{ BP } \) gives BP = 30        ....(2)
substituting (2) in (1). We get \(\sqrt { 3 } =\frac { 30+h }{ 30 } \) 
h = 30\((\sqrt { 3 } -1)\) = 30(1.732 - 1) = 30(0.732) = 21.96m.
Hence, the height of the tower is 21.96 m.

Let AB be the height of the kite above the ground. Then, AB = 75.
Let AC be the length of the string.
In right triangle ABC,\(\angle \)ACB = \(60°\)
\(sin\theta =\frac { AB }{ AC } \)
\(sin60°=\frac { 75 }{ AC } \)
gives \(\frac { \sqrt { 3 } }{ 2 } =\frac { 75 }{ AC } \) so, AC = \(\frac { 150 }{ \sqrt { 3 } } =50\sqrt { 3 } \) 
Hence, the length of the string is 50\(\sqrt { 3 } m\)

Let AB the lighthouse. Let C and D be the positions of the two ships.\(\times \)
Then, AB = 200m.
\(\angle ACB=30°,\angle ADB=45°\)
In right triangles BAC, tan30°= \(\frac { AB }{ Ac } \) 
\(\frac { 1 }{ \sqrt { 3 } } =\frac { 200 }{ AC } \ gives\ AC=200\sqrt { 3 } \)      ...(1)
In the right triangle BAD,tan45°= \(\frac { AB }{ AD } \)
\(1=\frac { 200 }{ AD } \) gives AD = 200     ...(2)
Now, CD = AC + AD = \(200\sqrt { 3 } +200\) [by(1) and (2)]
CD = 200\((\sqrt { 3 } +1)\) = 200 x 2.732 = 546.4
Distance between two ships is 546.4m

Given \(\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a} \)
\(\therefore a=\frac{1+\sin \theta}{\cos \theta} \)
\(\mathrm{LHS}=\frac{a^{2}-1}{a^{2}+1} \)
\(=\frac{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}-1}{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}+1}\)
\(=\frac{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}-1}{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}+1}\)
\(=\frac{\frac{1+\sin ^{2} \theta+2 \sin \theta-\cos ^{2} \theta}{\cos ^{2} \theta}}{\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{\cos ^{2} \theta}}\)
\(=\frac{\left(1-\cos ^{2} \theta\right)+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{1+\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+2 \sin \theta}\)
\(=\frac{\sin ^{2} \theta+\sin ^{2} \theta+2 \sin \theta}{1+1+2 \sin \theta}\)
\(=\frac{2 \sin ^{2} \theta+2 \sin \theta}{2+2 \sin \theta} \)
\(=\frac{2 \sin \theta(\sin \theta+1)}{2(1+\sin \theta)} \)
= sin \(\theta \) = RHS

sin \(\theta \) + cos \(\theta \) = P and sec \(\theta \) + cosec \(\theta \) = 9
LHS = q(p² - 1)
= (sec \(\theta \) + cosec \(\theta \)) [(sin \(\theta \) + cos \(\theta \))² -1]
\(=\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1\right) \)
\(=\left(\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\right)(1+2 \sin \theta \cos \theta-1) \)
\(=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta} \times 2 \sin \theta \cos \theta \)
\(=2(\sin \theta+\cos \theta)=2 p=\text { RHS } \)

Given
\( \frac{\cos \alpha}{\cos \beta}=m \)
\(\frac{\cos \alpha}{\sin \beta} =n \)
\(\text { LHS } =\left(m^{2}+n^{2}\right) \cos ^{2} \beta \)
\(=\left(\frac{\cos ^{2} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}\right) \cos ^{2} \beta \)
\(=\frac{\left(\cos ^{2} \alpha \sin ^{2} \beta+\cos ^{2} \alpha \cos ^{2} \beta\right)}{\cos ^{2} \beta \sin ^{2} \beta} \cos ^{2} \beta \)
\(=\frac{\cos ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right)}{\sin ^{2} \beta} \)
\(=\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}(1) \)
\(=\left(\frac{\cos \alpha}{\sin \beta}\right)^{2} \)
= n² = RHS

We have sin \(\theta \) + cos \(\theta \) = \(\sqrt{3}\)
Squaring on both the sides,
\((\sin \theta+\cos \theta)^{2} =(\sqrt{3}) \)
\(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta =3 \)
1 + 2 sin \(\theta \) cos \(\theta \) = 3
2 sin \(\theta \) cos \(\theta \) = 3 - 1
2 sin \(\theta \) cos \(\theta \) = 2
sin \(\theta \) cos \(\theta \) = 2/2
sin \(\theta \) cos \(\theta \) = 1
Now to prove tan \(\theta \) + cot \(\theta \) = 1
\( \mathrm{LHS} =\tan \theta+\cot \theta \)
\(=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta} \)
\(=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta} \)
\(=\frac{1}{1}\)
= 1
tan \(\theta \) + cot \(\theta \)  = 1

tan²A - tan²B = \(\frac { si{ n }^{ 2 }A-si{ n }^{ 2 }B }{ co{ s }^{ 2 }Aco{ s }^{ 2 }B } \) 
\(=\frac { si{ n }^{ 2 }Aco{ s }^{ 2 }B-si{ n }^{ 2 }Bco{ s }^{ 2 }A }{ co{ s }^{ 2 }Aco{ s }^{ 2 }B } \)
\(=\frac { si{ n }^{ 2 }A(1-si{ n }^{ 2 }B)-si{ n }^{ 2 }B(1-si{ n }^{ 2 }B(1-si{ n }^{ 2 }A) }{ co{ s }^{ 2 }Aco{ s }^{ 2 }B } \)
\(=\frac { si{ n }^{ 2 }A-si{ n }^{ 2 }Asi{ n }^{ 2 }B-si{ n }^{ 2 }B+si{ n }^{ 2 }Asi{ n }^{ 2 }B }{ co{ s }^{ 2 }Aco{ s }^{ 2 }B } =\frac { si{ n }^{ 2 }A-si{ n }^{ 2 }B }{ co{ s }^{ 2 }Aco{ s }^{ 2 }B } \)

Given cosec\(\theta \) + cot\(\theta \) = p   ...(1)
cosec²\(\theta \) - cot²\(\theta \) = 1 (identity)
\(\operatorname{cosec} \theta-\cot \theta=\frac{1}{\operatorname{cosec} \theta+\cot \theta}\)
cosec\(\theta \) - cot\(\theta \) =\(\frac { 1 }{ { p } } \)  .... (2)
Adding(1) and (2) we get, 2cosec\(\theta \) = \(p+\frac { 1 }{ p } \) 
2cosec\(\theta \)\(\frac { { p }^{ 2 }+1 }{ p } \) ....(3) 
Subtracting (2) from (1), we get, 2cot\(\theta \) = \(p-\frac { 1 }{ p } \) 
2cot\(\theta \) = \(\frac { { p }^{ 2 }-1 }{ p } \)   ...(4)
Dividing (4) by (3) we get,\(\frac { 2cot\theta }{ 2cosec\theta } =\frac { { p }^{ 2 }-1 }{ p } \times \frac { p }{ { p }^{ 2 }+1 } gives,cos\theta =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

\(\frac { sinA }{ 1+cosA } +\frac { sinA }{ 1-cosA } \)
\(=\frac { sinA(1-cosA)+sinA(1+cosA) }{ (1+cosA)(1-cosA) } \)
\(=\frac { sinA-sinAcosA+sinA+sinAcosA }{ 1-co{ s }^{ 2 }A } \)
\(=\frac { 2sinA }{ 1-co{ s }^{ 2 }A } =\frac { 2sinA }{ si{ n }^{ 2 }A } \) = 2cosecA

(cosec\(\theta \) - sin\(\theta \)) (sec\(\theta \) - cos\(\theta \)) (tan\(\theta \) + cot\(\theta \))
\(\left( \frac { 1 }{ sin\theta } -sin\theta \right) \quad \left( \frac { 1 }{ cos\theta } -cos\theta \right) \left( \frac { sin\theta }{ cos\theta } +\frac { cos\theta }{ sin\theta } \right) \)
\(=\frac { 1-si{ n }^{ 2 } }{ sin\theta } \times \frac { 1-co{ s }^{ 2 } }{ cos\theta } \times \frac { si{ n }^{ 2 }\theta +co{ s }^{ 2 }\theta }{ sin\theta cos\theta } \)
\(=\frac { co{ s }^{ 2 }\theta si{ n }^{ 2 }\theta \times 1 }{ si{ n }^{ 2 }\theta co{ s }^{ 2 }\theta } =1\)

Now,cos\(\theta \) + sin\(\theta \) =\(\sqrt { 2 } \) cos\(\theta \) 
Squaring both sides,
(cos\(\theta \) + sin\(\theta \) )² =(\(\sqrt { 2 } \) cos\(\theta \) )²
cos² \(\theta \) + sin² \(\theta \) + 2sin\(\theta \) cos\(\theta \)  = 2cos²\(\theta \) 
2cos²\(\theta \) - cos²\(\theta \) - sin²\(\theta \)  = 2sin\(\theta \) cos\(\theta \) 
cos²\(\theta \)  - sin²\(\theta \)  = 2sin\(\theta \) cos\(\theta \) 
(cos\(\theta \) + sin\(\theta \) ) (cos\(\theta \) + sin\(\theta \) ) = 2sin\(\theta \) cos\(\theta \) 
cos\(\theta \) - sin\(\theta \)  = \(\frac { 2sin\theta cos\theta }{ cos\theta +sin\theta } \) =\(\frac { 2sin\theta cos\theta }{ \sqrt { 2 } cos\theta } \) [since cos\(\theta \)  + sin\(\theta \) =\(\sqrt { 2 } \) cos\(\theta \) ]
=\(\sqrt { 2 } \) cos\(\theta \) 
Therefore cos\(\theta \) - sin\(\theta \)  =\(\sqrt { 2 } \) cos\(\theta \)

sin² Acos² B + cos² Asin² B + cos² A + cos² B+  sin² Asin² B
= sin² Acos² B + sin² Asin² B + cos² A + cos² B + sin²Asin² B
= sin² A(cos² B + sin² B) + cos² A(sin² B + cos²B)
= sin² A(1) + cos² A(1) (since sin² B + cos² B = 1) 
= sin² A + cos² A = 1