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 

(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) \(\frac { sin\ 49° }{ cos\ 41° } \) 
sin 49⁰ = sin(90⁰ - 41⁰) = cos 41⁰,  since 49⁰ + 41⁰ = 90⁰ (complementary),
Hence on substituting sin 49^o = cos41^o we get, \( \frac { cos\ 41° }{ cos\ 41° } \)= 1
(ii) \(\frac { sec\ 63° }{ cosec\ 27° } \)
sec63^o = sec (90^o- 27^o) = cosec27^o, here, 63^o and 27^o are complementary angles
we have \(\frac { sec\ 63° }{ cosec\ 27° } =\frac { cosec\ 27° }{ cosec\ 27° } =1\)

(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

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 = \(\sqrt { 13 } \)
\(sin\ A=\frac { opposite\ side }{ hypotenuse } =\frac { 2 }{ \sqrt { 13 } } \)
\(cosec\ A=\frac { hypotenuse }{ opposite\ side } =\frac { \sqrt { 13 } }{ 2 } \)
\(cos\ A=\frac { adjacent\ side }{ hypotenuse } =\frac { 3 }{ \sqrt { 13 } } \)
\(sec\ A=\frac { hypotenuse }{ adjacent\ side } =\frac { \sqrt { 13 } }{ 3 } \)
\(cot\ A=\frac { adjacent\ side }{ opposite\ side }= \frac { 3 }{ 2 } \)

By Pythagoras theorem,
\(AB=\sqrt { { BC }^{ 2 }{ AC }^{ 2 } } \)
= \(\sqrt { { \left( 25 \right) }^{ 2 }-{ 7 }^{ 2 } } \)
= \(\sqrt { 625-49 } =\sqrt { 576 } \) = 24
The six trignometric ratios are
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { 7 }{ 25 } \)
\(tan\theta =\frac { oppositeside }{ adjacent\ side } =\frac { 7 }{ 24 } \)
\(sec\theta =\frac { hypotenuse }{ adjacent\ side } =\frac { 25 }{ 24 } \)
\(cos\theta =\frac { adjacentside }{ hypotenuse } =\frac { 24 }{ 25 } \)
\(cosec\theta =\frac { hypotenuse }{ oppositeside } =\frac { 25 }{ 7 } \)
\(cot \theta\ \frac { adjacentside }{ oppositeside } =\frac { 24 }{ 7 } \)

In the given right angled triangle, note that for the given angle \(\theta \), PR is the ‘opposite’ side and PQ is the ‘adjacent’ side.
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { PR }{ QR } =\frac { 35 }{ 37 } \)
\(cos\theta =\frac { adjacent\ side }{ hypotenuse } =\frac { PQ }{ QR } =\frac { 12 }{ 37 } \)
\(tan\theta =\frac { opposite\ side }{ adjacent\ side } =\frac { PR }{ PQ } =\frac { 35 }{ 12 } \)
It is enough to leave the ratios as fractions. In case, if you want to simplify each ratio neatly in a terminating decimal form, you may opt for it, but that is not obligatory.

\(tan\ {42 }^{ 0 }=\cfrac { h }{ 60 } =0.9004\)
h = 0.9004 \(\times\) 60 = 54.024 m

\(cos\theta =\cfrac { 4 }{ 5 } =0.8\)
cos 36° 48' = 0.8
\(\therefore\) \(\theta\) = 36⁰48'

Hypotenuse = 10 cm
One of the acute angle = 24° 24'
sin 24° 24' = 0.4131'
\(\cfrac { x }{ 10 } =0.4131\)
x = 0.4131 \(\times\) 10
x = 4.131
cos 24° 24' = 0.9107
\(\cfrac { y }{ 10 } =0.9107\)
y = 9.107
\(\therefore\) Area of the triangle = \(\cfrac { 1 }{ 2 } bh\)
= \(\cfrac { 1 }{ 2 } \times y\times x\)
\(=\cfrac { 1 }{ 2 } \times 9.107\times 4.131=18.81sq.cm\)

(i) = 0.9111 + 0.9066 + 0.1793
= 1.9970
(ii) = 2.8982 + 0.9639 - 0.9962
= 3.8625 - 0.9962
= 2.8659

(i) From the natural sines table
sin 85° 57' = 0.9975
\(\therefore\) \(\theta \) = 85° 57'
(ii) cos\(\theta \) = 0.6763
cos 47° 33' = 0.6762
\(\therefore\) \(\theta \) = 47° 33'
(iii) tan \(\theta \) = 0.0720
tan 4° 7' = 0.0720
\(\therefore\)\(\theta \) = 4°7'
(iv) cos \(\theta \) = 0.0410
cos 87° 45° = 0.0410
:\(\therefore\)\(\theta \) = 87°39'
(v) tan \(\theta \) = 7.5958
tan 82° 30' = 7.5958
\(\therefore\) \(\theta \) = 82° 30'

(i) sin 49° = 0.7547
(ii) cos 74° 39'
cos 74° 36' = 0.2656 (From the natural cosines table)
Mean difference 3' = 8
cos 74° 39'= 0.2648
(iii) tan 54° 26'
From the natural tangents table
tan 54° 24' = 1.3968
Mean difference 2' = 17
tan 54° 26' = 1.3985
(iv) sin 21°21'
From the natural sines table
sin 21° 18' = 0.3633
Mean difference 3' = 8
sin 21° 18' = 0.3641
(v) cos 33° 53'
From the natural cosines table
cos 33° 48' = 0. 8310
Mean difference 5' = 8
cos 33° 53' = 0.8318
(vi) tan70^o12' = 2.7776
Mean difference for 5'= 131 (Mean diference is to be added)
= 2.7907

8 sin 2(15°). cos 4 (15°). sin 6(15°)
= 8 sin 30° cos 60° sin 90°
= \(8\times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times 1=2\)

L.H.S = cos 3A= cos 3 (30°)
= cos 90°
=0 ---- (1)
R.H.S = 4 cos³ A- 3 cosA
= 4 cos³ 30° - 3 cos 30°
= \(4\left( \cfrac { \sqrt { 3 } }{ 2 } \right) ^{ 3 }-3\times \cfrac { \sqrt { 3 } }{ 2 } \)

(1) = (2). Hence it is verified

(i) \(\frac { tan45° }{ cosec30° } +\frac { sec60° }{ cot45° } -\frac { 5sin90° }{ 2cos0° } \)
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 1 } -\cfrac { 5\times 1 }{ 2\times 1 } \)
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 1 } -\cfrac { 5\times 1 }{ 2\times 1 } \)
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 1 } -\cfrac { 5 }{ 2 } =\cfrac { 1+4-5 }{ 2 } =\cfrac { 5-5 }{ 2 } =\cfrac { 0 }{ 2 } =0\)
(ii) (sin 90° + cos 60° + cos 45°) \(\times\) (sin 30° - cos 0° + cos 45⁰)
= \(\left( 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ \sqrt { 2 } } \right) \times \left( \cfrac { 1 }{ 2 } -1+\cfrac { 1 }{ \sqrt { 2 } } \right) \)
= \(\cfrac { 2+1+\sqrt { 2 } }{ 2 } \times \cfrac { 1-2+\sqrt { 2 } }{ 2 } \)
= \(\cfrac { 3+\sqrt { 2 } }{ 2 } \times \cfrac { \sqrt { 2 } -1 }{ 2 } =\cfrac { 3\sqrt { 2 } +2-3-\sqrt { 2 } }{ 4 } \)
= \(\cfrac { 3\sqrt { 2 } -1-\sqrt { 2 } }{ 4 } =\cfrac { 2\sqrt { 2 } -1 }{ 4 } \) \(=\frac{18-4}{8}=\frac{7}{4}\)
(iii) sin²30⁰ - 2 cos³60⁰ + 3 tan⁴45⁰

 
\({ \sin }^{ 2 }{ \tan }^{ 2 }+{ \cos }^{ 2 }{ 60 }^{ o }=\left( \cfrac { \sqrt { 3 } }{ 2 } \right) +\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }=\cfrac { 3 }{ 4 } +\cfrac { 1 }{ 4 } =\cfrac { 4 }{ 4 } =1\)

In the figure,
\(\triangle\)OPM, \(\triangle\)OQN are similar triangles. In similar triangles the sides are in the same proportional.
\(\cfrac { QN }{ PM } =\cfrac { QO }{ PO } \)
\(\cfrac { h }{ 5 } =\cfrac { 35 }{ 25 } \)
\(h=\cfrac { 5\times 35 }{ 25 } \)

h = 7m

In\(\triangle\)ABC
AC² = 15²= 225- - - - - (1)
BC² = 20² = 400 - - - - (2)
AB²= (9 + 16)²
From (1), (2), (3)
AB² = AC² + BC²
625 = 225 + 400 = 625
\(\therefore \angle C=\theta +\Phi ={ 90 }^{ 0 }\)
( \(\therefore\) By Pythagoras theorem, in a right angled triangle square of hypotenuse is equal to sum of the squares of other two side)
And also in the figure. \(\triangle\)ADC, \(\triangle\)DBC are two other triangles.
As per the data given,
9²+ 12²= 81 + 144 = 225 = 15²
\(\therefore\) \(\triangle\)ADC is a right angled triangle.
then 12²+ 16² = 144 + 256 = 400 = 20²
\(\therefore\) \(\triangle\)DBC is also a right angled triangle.
\(sin\alpha =\cfrac { 12 }{ 15 } =\cfrac { 4 }{ 5 } ,cos\beta =\cfrac { 16 }{ 20 } =\cfrac { 4 }{ 5 } ,tan\phi =\cfrac { 16 }{ 12 } =\cfrac { 4 }{ 3 } \)

\(cos\theta :sin\theta =1:2\)
\(\cfrac { cos\theta }{ sin\theta } =\cfrac { 1 }{ 2 } \)
\(cos\theta =\cfrac { 1 }{ 2 } sin\theta \)
\(\sin\theta =2 \cos\theta \)

\(\therefore \cfrac { 8cos\theta -2sin\theta }{ 4cos\theta +2sin\theta } =\cfrac { 1 }{ 2 } \)

\(sin\ \theta =\cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(cos\ \theta =\cfrac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(b\ sin\ \theta =b\times \cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } -\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ... (1)
\(a\ cos\ \theta =a\times \cfrac { b }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } } =\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ...(2)
(1) = (2)\(\Rightarrow\) LHS = RHS
Hence proved

By By the Pythagoras theorem
BC² = AC² + AB²
= \(\left( \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \right) ^{ 2 }-{ a }^{ 2 }\)
= a²+b²-a²
= b²
BC = b

By the pythagoras theorem,
AB² = OA² + OB²
(1 + x²)² = (2x)² + OB²
OB² = (1 +x²)² - (2x)² = 1+ x⁴ + 2x² - 4x² = 1+x⁴ - 2x²
OB² = (1-x²)² B
OB = (1-x²)
\(\therefore sin\ A=\cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(tan\ A=\cfrac { 1-{ x }^{ 2 } }{ 2x } \)

\(sinA=\cfrac { 4 }{ 5 } \)
\(tanA=\cfrac { 4 }{ 3 } \)
\(\therefore \cfrac { sinA-cosA }{ 2tanA } =\cfrac { \frac { 4 }{ 5 } -\frac { 3 }{ 5 } }{ 2\times \frac { 4 }{ 3 } } =\cfrac { \frac { 1 }{ 5 } }{ 2\times \frac { 4 }{ 3 } } =\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 5 } \times \cfrac { 3 }{ 4 } =\cfrac { 3 }{ 40 } \)

By the Pythagoras theorem
x=\(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } \)
= \(\sqrt { 25-9 } \)
= \(\sqrt { 16 } =4\)

If \(2cos\theta =\sqrt { 3 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(x=\sqrt { { 2 }^{ 2 }-\sqrt { { 3 }^{ 2 } } } =\sqrt { 4-3 } =\sqrt { 1 } =1\)
\(\therefore sin\theta =\cfrac { 1 }{ 2 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(tan\theta =\cfrac { 1 }{ \sqrt { 3 } } \)
\(coseec\theta =2\)
\(sec\theta =\cfrac { 2 }{ \sqrt { 3 } } \)
\(cot\theta =\sqrt { 3 } \)

(i) \(sinB=\cfrac { 12 }{ 13 } \)
(ii) \(secB=\cfrac { 1 }{ cosB } \)
=  \(\cfrac { 1 }{ 5/3 } =\cfrac { 13 }{ 5 } \)
(iii) \(cotB=\cfrac { 1 }{ tanB } =\cfrac { 1 }{ 12/5 } =\cfrac { 5 }{ 2 } \)
(iv) 

(v) \(\tan C=\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } \)
(vi)  \(cosecC=\cfrac { 1 }{ sinC } =\cfrac { 1 }{ 12/20 } =\cfrac { 20 }{ 12 } =\cfrac { 5 }{ 3 } \)

By the pythagoras theorem,
\(AD=\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \)
= \(\sqrt { 144 } =12\)
AC =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\) 

By the Pythagoras theorem
In \(\triangle\)OAB, \(x=\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } } =\sqrt { 100-64 } =\sqrt { 36 } \)
\(sin\theta =\cfrac { 8 }{ 10 } =\cfrac { 4 }{ 5 } \)
\(cos\theta =\cfrac { 6 }{ 10 } =\cfrac { 3 }{ 5 } \)
\(tan\theta =\cfrac { 8 }{ 6 } =\cfrac { 4 }{ 3 } \)
\(cosec\theta =\cfrac { 10 }{ 8 } =\cfrac { 5 }{ 4 } \)
\(sec\theta =\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } \)
\(cot\theta =\cfrac { 6 }{ 8 } =\cfrac { 3 }{ 4 } \)
Study these thoroughly
\(sin\theta =\cfrac { Opp.side }{ Hypotenuse } \)
\(cos\theta =\cfrac { Adj.side }{ Hypotenuse } \)
\(tan\theta =\cfrac { Opp.side }{ Adj.side } \)
\(cosec\theta =\cfrac { Hypotenuse }{ Opp.side } \)
\(sec\theta =\cfrac { Hypotenuse }{ Opp.side } \)
\(cot\theta =\cfrac { Adj.side }{ Opp.side } \)

sin B = \(\frac { 9 }{ 41 } \);
cos B = \(\frac { 40 }{ 41 } \);
tan B =\(\frac { 9 }{ 40 } \) ;
cosec B = \(\frac { 1 }{ sinB } \) = \(\frac { 41 }{ 9 } \);
sec B =\(\frac { 1 }{ cotB } \) = \(\frac { 41 }{ 40 } \);
cot B =\(\frac { 1 }{ tanB } \) = \(\frac { 40 }{ 9 } \)

(i) sin \(\theta\) = 0.9858  = 0.9857 + 0.0001
From the sine table 0.9857 = 80^o 18'
Mean difference 1 = 2′
0.9858 = sin 80⁰20'
sin\(\theta\) = 0.9858 = sin 80⁰20'
\(\theta\) = 80⁰20'
(ii) cos \(\theta\) = 0.7656 =  0.7660 - 0.0004
From the natural cosine table
0.7660 = 40°0′
Mean difference 4 = 2′       
0.7656 = 40°0′         
cos \(\theta\) = 0.7656 = cos 40°2'
\(\theta\) = 40°2'                                                                                          

(i) (cos0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
= \(\left[ 1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \left[ 1-\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \)
= \(\left[ \frac { 2\sqrt { 2 } +2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] \left[ \frac { 2\sqrt { 2 } -2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] =\left[ \frac { 3\sqrt { 2 } +2 }{ 2\sqrt { 2 } } \right] \left[ \frac { 3\sqrt { 2 } -2 }{ 2\sqrt { 2 } } \right] \)
= \(\frac { 18-4 }{ 4\left( \sqrt { 2 } \right) ^{ 2 } } =\frac { 14 }{ 4\times 2 } =\frac { 7 }{ 4 } \) 
(ii) tan²60⁰ - 2tan²45⁰ - cot²30⁰ +2sin²30⁰ + \(\frac { 3 }{ 4 } \) cosec²45⁰
= \(\left( \sqrt { 3 } \right) ^{ 2 }-2(1)^{ 2 }-\left( \sqrt { 3 } \right) ^{ 2 }+2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 3 }{ 4 } \left( \sqrt { 2 } \right) ^{ 2 }\)
= \(3-2-3+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \)
= -2 + \(\frac { 4 }{ 2 } \) = -2 + 2 = 0