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

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

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

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 = \(\cfrac { 1 }{ 2 } bh\)
= \(\cfrac { 1 }{ 2 } \times y\times x\)
\(=\cfrac { 1 }{ 2 } \times 9.107\times 4.131=18.81sq.cm\)

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

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

In\(\triangle\)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.
\(sin\alpha =\cfrac { 12 }{ 15 } =\cfrac { 4 }{ 5 } ,cos\beta =\cfrac { 16 }{ 20 } =\cfrac { 4 }{ 5 } ,tan\phi =\cfrac { 16 }{ 12 } =\cfrac { 4 }{ 3 } \)

\(cot\ A=\cfrac { 2 }{ 3 } \)
\(cot\ A=\cfrac { Adjacent\ side }{ Opposite\ side } \)
\(tan\ A=\ \cfrac { Opp.side }{ Adj.side } =\cfrac { 3 }{ 2 } \)
\(sin\ A=\cfrac { 3 }{ \sqrt { 13 } } \)
\(cotA=\cfrac { 2 }{ \sqrt { 13 } } \)
\(\therefore \cfrac { 4sinA-3cosA }{ 2ainA+3cosA } =\cfrac { 4\times \cfrac { 3 }{ \sqrt { 13 } } -3\times \cfrac { 2 }{ \sqrt { 13 } } }{ 2\times \cfrac { 3 }{ \sqrt { 13 } } +3\cfrac { 2 }{ \sqrt { 13 } } } \)
= \(\cfrac { \frac { 12 }{ \sqrt { 13 } } -\frac { 6 }{ \sqrt { 13 } } }{ \frac { 6 }{ \sqrt { 13 } } +\frac { 6 }{ \sqrt { 13 } } } =\cfrac { \frac { 6 }{ \sqrt { 13 } } }{ \frac { 2 }{ \sqrt { 13 } } } =\cfrac { 6 }{ \sqrt { 13 } } \times \cfrac { \sqrt { 13 } }{ 12 } \)
= \(\cfrac { 6 }{ 12 } =\cfrac { 1 }{ 2 } \)

By Pythagoras theorem In\(\triangle\)OAB,
AB² = OA²+ OB²
= 3² + 2²
= 9+4
= 13
\(AB=\sqrt { 13 } \)

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)
\(\therefore sin\ A=\cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(tan\ A=\cfrac { 1-{ x }^{ 2 } }{ 2x } \)

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 =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\) 

(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