Question

Trigonometry Special Right Triangles Find the exact value of cosec 60* + sec 30* + sin 60*.

Answer 1

Answer:

11sqrt(3)/6

Step-by-step explanation:

sin(60) = sqrt(3)/2

cosec(60) = 2/sqrt(3) = 2sqrt(3)/3

cos(30) = sqrt(3)/2

sec(30) = 2sqrt(3)/3

2sqrt(3)/3 + 2sqrt(3)/3 + sqrt(3)/2

Lcm: 6

sqrt(3)(4+4+3)/6

11sqrt(3)/6

Answer 2

Answer:

= 3.175

Step-by-step explanation:

$\csc( \alpha ) = \frac{1}{ \sin( \alpha ) } \\ \sec( \alpha ) = \frac{1}{ \cos( \alpha ) } \\ \\ \\ \csc(60) + \sec(30) + \sin(60) \\ = \frac{1}{ \sin(60) } + \frac{1}{ \cos(30) } + \sin(60) \\ = \frac{1}{ \frac{ \sqrt{3} }{2} } + \frac{1}{ \frac{ \sqrt{3} }{2} } + \frac{ \sqrt{3} }{2} \\ = \frac{2}{ \sqrt{3} } + \frac{2}{ \sqrt{3} } + \frac{ \sqrt{3} }{2} \\ = \frac{4}{ \sqrt{3} } + \frac{ \sqrt{3} }{2} \\ = \frac{2(4) + ( \sqrt{3} ) \sqrt{3} }{2 \sqrt{3} } \\ = \frac{8 + 3}{2 \sqrt{3} } \\ = \frac{11}{2 \sqrt{3} } \\ = \frac{11}{2 \sqrt{3} } \times \frac{2 \sqrt{3} }{2 \sqrt{3} } \\ = \frac{22 \sqrt{3} }{4( \sqrt{3} )^{2} } \\ = \frac{22 \sqrt{3} }{4(3)} \\ = \frac{22 \sqrt{3} }{12} \\ = \frac{11 \sqrt{3} }{6} \\ = 3.175$