Using trigonometric tables- find the value of sin 64- 42-cos 42- 20-

The correct option is B 1.6433 We have sin64^∘42' = 0.9041 42^∘20' = 42^∘18' + 2' Now, cos 42^∘18' = 0.7396 Mean difference for 2’ = 0.0004 [To be subtracted] ...

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Standard X

Mathematics

Trigonometric Ratios of a Right Triangle

Using trigono...

Question

Using trigonometric tables, find the value of $(s i n 64^{\circ} 42^{'} + c o s 42^{\circ} 20^{'})$

1.5623

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1.6433

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1.234

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1.986

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Solution

The correct option is B
1.6433

We have

$s i n 64^{\circ} 42^{'} = 0.9041$

$42^{\circ} 20^{'} = 42^{\circ} 18^{'} + 2^{'}$

Now, $c o s 42^{\circ} 18^{'} = 0.7396$

Mean difference for 2’ = 0.0004 [To be subtracted]

So, $c o s 42^{\circ} 20^{'} = 0.7392$

Hence, $s i n 64^{\circ} 42^{'} + c o s 42^{\circ} 20^{'} = 0.9041 + 0.7392 = 1.6433$

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Using trigonometric tables, find the value of (sin64∘42′ + cos42∘20′)

The correct option is B 1.6433 We have sin64∘42′ = 0.9041 42∘20′ = 42∘18′ + 2′ Now, cos42∘18′ = 0.7396 Mean difference for 2’ = 0.0004 [To be subtracted] So, cos42∘20′ = 0.7392 Hence, sin64∘42′ + cos42∘20′ = 0.9041 + 0.7392 = 1.6433

Using trigonometric tables, find the value of $(s i n 64^{\circ} 42^{'} + c o s 42^{\circ} 20^{'})$