The value of tan49°50' is .

1.1268

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1.1847

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1.5112

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Solution

The correct option is A 1.1268
We can write,
49°50' = 49°48' + 2'
From the table we have, tan49°48' = 1.1833
Mean difference for 2′ = 14 (Mean difference to be added for tan)
tan49°48' = 1.1833
2′ = + 14
____________________
tan49°50' = 1.1847

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Q. Find

A

tan 49^{\circ} = cot A

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

(i) sin67° + cos75°
(ii) cot65° + tan49°
(iii) sec78° + cosec56°
(iv) cosec54° + sin72°

Express each of the following in terms of T-ratios of angles lying between $0^{\circ} a n d 45^{\circ} :$

(i) $s i n 67^{\circ} + c o s 75^{\circ}$ (ii) $c o t 65^{\circ} + t a n 49^{\circ}$

(iii) $s e c 78^{\circ} + c o s e c 56^{\circ}$ (iv) $c o s e c 54^{\circ} + s i n 72^{\circ}$

Q. The value of

{(\frac{50}{0})}^{2} + {(\frac{50}{1})}^{2} + {(\frac{50}{2})}^{2} + . . . + {(\frac{50}{49})}^{2} + {(\frac{50}{50})}^{2}

Q. The mean value of numbers

^{50} C_{0},^{50} C_{2},^{50} C_{4}, \dots^{50} C_{50}

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The value of tan49°50' is .

The correct option is A 1.1268We can write, 49°50' = 49°48' + 2' From the table we have, tan49°48' = 1.1833 Mean difference for 2′ = 14 (Mean difference to be added for tan) tan49°48' = 1.1833 2′ = + 14 ____________________ tan49°50' = 1.1847

The correct option is A 1.1268
We can write,
49°50' = 49°48' + 2'
From the table we have, tan49°48' = 1.1833
Mean difference for 2′ = 14 (Mean difference to be added for tan)
tan49°48' = 1.1833
2′ = + 14
____________________
tan49°50' = 1.1847