Question

ம‌தி‌ப்பு கா‌ண்க

sin38°〖36〗^'+ tan12°〖12〗^' tan60°〖25〗^'-cos49°〖20〗^'

Answer 1

i) 0.8401   ii) 1.1098

விளக்கம்:

i)$\sin 38^{\circ} 36^{\prime}+\tan 12^{\circ} 12^{\prime}$

$\sin 38^{\circ} 36^{\prime}=0.6239$

$\tan 12^{\circ} 12^{\prime}=0.2162$

$\sin 38^{\circ} 36^{\prime}+\tan 12^{\circ} 12^{\prime}$= 0.6239 + 0.2162

                                  = 0.8401

ii) $\tan 60^{\circ} 25^{\prime}-\cos 49^{\circ} 20^{\prime}$

$\tan 60^{\circ} 25^{\circ}=1.7603$ + 0.0012

                =  1.7615

$\cos 49^{\circ} 20^{\prime}=0.6521$ - 0.0004

               = 0.6517

                = 1.7615 - 0.6517

$\tan 60^{\circ} 25^{\prime}-\cos 49^{\circ} 20^{\prime}=1.1098$

Answer 2

Step-by-step explanation:

0.8401 ii) 1.1098

Solution

i)\sin 38^{\circ} 36^{\prime}+\tan 12^{\circ} 12^{\prime}sin38

∘

36

′

+tan12

∘

12

′

\sin 38^{\circ} 36^{\prime}=0.6239sin38

∘

36

′

=0.6239

\tan 12^{\circ} 12^{\prime}=0.2162tan12

∘

12

′

=0.2162

\sin 38^{\circ} 36^{\prime}+\tan 12^{\circ} 12^{\prime}sin38

∘

36

′

+tan12

∘

12

′

= 0.6239 + 0.2162

= 0.8401

ii) \tan 60^{\circ} 25^{\prime}-\cos 49^{\circ} 20^{\prime}tan60

∘

25

′

−cos49

∘

20

′

\tan 60^{\circ} 25^{\circ}=1.7603tan60

∘

25

∘

=1.7603 + 0.0012

= 1.7615

\cos 49^{\circ} 20^{\prime}=0.6521cos49

∘

20

′

=0.6521 - 0.0004

= 0.6517

= 1.7615 - 0.6517

\tan 60^{\circ} 25^{\prime}-\cos 49^{\circ} 20^{\prime}=1.1098tan60

∘

25

′

−cos49

∘

20

′

=1.1098