sin 30° cos 60° + cos 30° sin 60° = sin 90°
\(sin\ { 30 }^{ 0 }=\cfrac { 1 }{ 2 },sin\ { 60 }^{ 0 }=\cfrac { \sqrt { 3 } }{ 2 } \)
\(cos\ { 30 }^{ 0 }=\cfrac { \sqrt { 3 } }{ 2 } ,cos\ { 60 }^{ 0 }=\cfrac { 1 }{ 2 } \)
sin 90° = 1
 \(\therefore\) sin 30° cos 60°=\(\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } =\cfrac { 1 }{ 4 } \)
cos30° sin60° = \(\cfrac { \sqrt { 3 } }{ 2 } \times \cfrac { \sqrt { 3 } }{ 2 } =\cfrac { 3 }{ 4 } \)
\(\therefore\) sin 30° cos 60° + cos 30° sin 60°
= \(\cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } =\cfrac { 4 }{ 4 } =1={ sin }^{ 0 }{ 90 }^{ 0 }\)

Study the above table thoroughly.

cos 90° = 1 - 2 sin² 45° = 2 cos²45° - 1 .
cos 90° = 0 - - - - - (1)

\(2cos^{ 2 }=2\times \left( \cfrac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }=2\times \cfrac { 1 }{ 2 } \)
2 cos² 45° - 1 = 1 - 1 = 0
(1) = (2) = (3). Hence it is verified.

1+ tan² 30° = sec 30°
1 + tan² 30° = \(1+\left( \cfrac { 1 }{ \sqrt { 3 } } \right) ^{ 2 }=1+\cfrac { 1 }{ 3 } =\cfrac { 3+1 }{ 3 } +\cfrac { 4 }{ 3 } \)
\({ sec }\ 30^{ 0 }=\cfrac { 1 }{ cos30 } =\cfrac { 1 }{ \sqrt { 3 } } =1\times \cfrac { 2 }{ \sqrt { 3 } } =\cfrac { 2 }{ \sqrt { 3 } } \)
\({ sec }^{ 2 }30=\left( \cfrac { 2 }{ \sqrt { 3 } } \right) ^{ 2 }=\cfrac { 4 }{ 3 } \)
(1) = (2) \(\Rightarrow\) LHS = RHS
\(\therefore\) Hence it is verified

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

\(\left( \frac { cos47° }{ sin43° } \right) +\left( \frac { sin72° }{ cos18° } \right) -2cos^{ 2 }45°\)
= \(\left( \cfrac { cos\left( { 90 }^{ 0 }-{ 43 }^{ 0 } \right) }{ { sin43 }^{ 0 } } \right) ^{ 2 }+\left( \cfrac { sin\left( { 90 }^{ 0 }-{ 18 }^{ 0 } \right) }{ cos18 } \right) ^{ 2 }-2{ cos }^{ 0 }{ 45 }^{ 0 }\)
= \(\left( \cfrac { sin43 }{ sin43 } \right) ^{ 2 }+\left( \cfrac { cos18 }{ cos18 } \right) ^{ 2 }-2\left( \cfrac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }\)