Question

Using trigonometric tables, find the value of $\frac{sin{64}^{\circ }{42}^{\prime }+cos{42}^{\circ }{20}^{\prime }}{tan{36}^{\circ }{40}^{\prime }+cot{63}^{\circ }{20}^{\prime }}$

• A. 1.31
• B. 1.89
• C. 1.013
• D. 1.80

Answer 1

The correct option is A

1.31

For numerator we have

$sin{64}^{\circ }{42}^{\prime }=0.9041$

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

Now, $cos{42}^{\circ }{18}^{\prime }=0.7396$

Mean difference for $2^{\prime }=0.0004$ [To be subtracted]

So, $cos{42}^{\circ }{18}^{\prime }=0.7392$

Hence, $sin{64}^{\circ }{42}^{\prime }+cos{42}^{\circ }{20}^{\prime }=0.9041+0.7396=1.6433$

For denominator we have,

${36}^{\circ }{40}^{\prime }={36}^{\circ }{36}^{\prime }+4^{\prime }$

$tan{36}^{\circ }{36}^{\prime }=0.7427$

Mean difference for $4^{\prime }=0.0018$

From trigonometric table, we have $tan{36}^{\circ }{40}^{\prime }=0.7445$

${63}^{\circ }{20}^{\prime }={63}^{\circ }{18}^{\prime }+2^{\prime }$

Now, $cot{63}^{\circ }{18}^{\prime }=0.5029$

Mean difference for 2’ = 0.0007 [To be subtracted]

$cot{63}^{\circ }{20}^{\prime }=0.5022$

So, $tan{36}^{\circ }{40}^{\prime }+cot{63}^{\circ }{20}^{\prime }=1.2467$

$\frac{sin{64}^{\circ }{42}^{\prime }+cos{42}^{\circ }{20}^{\prime }}{tan{36}^{\circ }{40}^{\prime }+cot{63}^{\circ }{20}^{\prime }}$ = $\frac{1.6433}{1.2467}$ = 1.318