Question

Without using trigonometric tables, evaluate the following:

`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`

Answer 1

(i) We have,

`\frac{\cos37^\text{o}}{\sin53^\text{o}}=\frac{\cos(90^\text{o}-53^\text{o})}{\sin53^\text{o}}=\frac{\sin53^\text{o}}{\sin53^\text{o}}=1`

[∵ cos(90º – θ) = sin θ]

(ii) We have,

`\frac{\sin 41^\text{o}}{\cos 49^\text{o}}=\frac{\sin(90^\text{o}-49^\text{o})}{\cos 49^\text{o}}=\frac{\cos49^\text{o}}{\cos 49^\text{o}}=1`

[∵ sin (90º – θ) = cos θ]

(iii) We have,

`\frac{\sin30^\text{o}17'}{\cos59^\text{o}43'}=\frac{\sin(90^\text{o}-59^\text{o}43')}{\cos59^\text{o}43'}=\frac{\cos59^\text{o}43'}{\cos 59^\text{o}43'}=1`