From the figure,
sin \(\theta\) = \(\frac { AB }{ AC } \)
sin 48⁰30' = \(\frac { AB }{ 5 } \)
0.7490 = \(\frac { AB }{ 5 } \)
5 \(\times\) 0.7490 = AB
AB = 3.7450 cm
cos \(\theta\) = \(\frac { BC }{ AC } \)
cos 48⁰30' = \(\frac { BC }{ 5 } \)
0.6626 = \(\frac { BC }{ 5 } \)
0.6626 \(\times\) 5 = BC
BC  = 3.313 cm
Area of right triangle = \(\frac { 1 }{ 2 } \) bh
\( =\frac{1}{2} \times B C \times A B \)
\( =\frac{1}{2} \times 3.3130 \times 3.7450 \)
\( =1.6565 \times 3.7450 \) = 6.2035925 cm²

(i) sin 38⁰36' + tan 12⁰12'
sin 38⁰36' = 0.6239
tan12⁰12' = 0.2162
sin 38⁰36' + tan 12⁰12' = 0.8401
(ii) tan 60⁰25^' - cos 49⁰20^'
tan 60⁰25' = 1.7603 + 0.0012 = 1.7615
cos 49⁰20^' = 0.6521 - 0.0004 = 0.6517
tan 60⁰25' - cos 49⁰20' = 1.1098

write 70⁰13' = 70⁰12' + 1'
From the table we have, tan70⁰12' = 2.7776 

write 19⁰59' = 19⁰54' + 5'
From the table we have, cos19⁰54' = 0.9403 

write 64⁰34' = 64⁰30' + 4'
From the table we have, sin64⁰30' = 0.9026 

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

(i) tan 7°tan 23°tan 60°tan 67°tan 83°
=  tan 7⁰ tan 83⁰ tan 23⁰ tan 67⁰ tan 60⁰ (Grouping complementary angles)
=  tan 7⁰ tan(90⁰ - 7⁰)tan 23⁰ tan(90⁰ - 23⁰)tan 60⁰
=  (tan7⁰.cot7⁰)(tan 23⁰. cot 23⁰)tan 60⁰
=  (1)\(\times\) (1)\(\times\) tan 60⁰
=  tan 60⁰ = \(\sqrt { 3 } \) 
(ii) \(\frac { cos35° }{ sin55° } +\frac { sin12° }{ cos78° } -\frac { cos18° }{ sin72° } \)
\(=\frac { cos\left( 90°-55° \right) }{ sin55° } +\frac { sin\left( 90°-78° \right) }{ cos78° } -\frac { cos\left( 90°-72° \right) }{ sin72° } \) \(\left[ \begin{matrix} { \text Since} \\cos35°=cos\left( 90°-55° \right) \\ sin12°=sin(90°-78°) \\ cos18°=cos\left( 90°-72° \right) \end{matrix} \right] \)
= \(\frac { sin55° }{ sin55° } +\frac { cos78° }{ cos78° } -\frac { sin72° }{ sin72° } \)
= 1+1 - 1 = 1

(i) sin74° = sin(90⁰ -16⁰)    (since, 90⁰ -16⁰ = 74⁰ )
RHS is of the form sin(90⁰ - \(\theta\)) = cos\(\theta\)
Therefore sin74° =  cos16⁰
(ii)  tan12° =  tan(90⁰ - 78⁰) (since, 12⁰= 90⁰ = 78⁰ )
RHS is of the form tan(90⁰-  \(\theta\)) = cot \(\theta\)
Therefore tan12° = cot 78⁰
(iii) cosec 39° = cosec(90⁰ -  51⁰) (since, 39⁰ = 90⁰ - 51⁰ )
RHS is of the form cosec(90⁰ - \(\theta\)) = sec\(\theta\)
Therefore cosec39° = sec51⁰

Let BC = 13 and AB = 5
sec θ = \(\frac { hypotenuse }{ adjacentside } =\frac { BC }{ AB } =\frac { 13 }{ 5 } \)
By the Pythagoras theorem,
\(AC=\sqrt { { BC }^{ 2 }-{ AB }^{ 2 } } \)
= \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \) = \(\sqrt { 144 } \) = 12
Therefore, \(sin\theta =\frac { AC }{ BC } =\frac { 12 }{ 13 } \) ; \(cos\theta =\frac { AB }{ BC } =\frac { 5 }{ 13 } \)
\(LHS=\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta } =\frac { 2\times \frac { 12 }{ 13 } =3\times \frac { 5 }{ 13 } }{ 4\times \frac { 12 }{ 13 } -9\times \frac { 5 }{ 13 } } =\frac { \frac { 24-15 }{ 13 } }{ \frac { 48-45 }{ 12 } } =\frac { 9 }{ 3 } =3\) = RHS

tan A = \(\frac { opposite\ side }{ adjacent\ side } =\frac { 2 }{ 3 } \)
By Pythagoras theorem,
\(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
= \(\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } \)
AC = \(\sqrt { 13 } \)
\(sin\ A=\frac { opposite\ side }{ hypotenuse } =\frac { 2 }{ \sqrt { 13 } } \)
\(cosec\ A=\frac { hypotenuse }{ opposite\ side } =\frac { \sqrt { 13 } }{ 2 } \)
\(cos\ A=\frac { adjacent\ side }{ hypotenuse } =\frac { 3 }{ \sqrt { 13 } } \)
\(sec\ A=\frac { hypotenuse }{ adjacent\ side } =\frac { \sqrt { 13 } }{ 3 } \)
\(cot\ A=\frac { adjacent\ side }{ opposite\ side }= \frac { 3 }{ 2 } \)