Solving for Angle 'a' in Quadrant 2: Math Homework

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SUMMARY

The problem involves solving for angle 'a' in the second quadrant given that (csc a) = (sec 0.75). The solution process begins with the identity 1/sin(a) = 1/cos(0.75), leading to the calculation of cos(0.75) = 0.7317. From this, sin(a) is determined to be 0.7317, yielding the principal root a = 0.8208 radians. Since 'a' is in the second quadrant, the final measure is calculated as a = π - 0.8208, resulting in a = 2.3208 radians.

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Homework Statement



Given that (csc a) = (sec0.75) and that 'a' lies in the second quadrant, determine the measure of angle 'a', to two decimal places.

Homework Equations



reciprocal and quotient identities.

The Attempt at a Solution



I only got as far as:

1/sin(a) = 1/cos(0.75)

The answer in the back of the book is 2.32

Can anyone guide me towards the light? thanks in advance!
 
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Well, it is not difficult to find that cos(.75)= .7317 using a calculator and , again using a calculator, that sin(a)= .7317 for a= .8208. Of course, that is the "principal root", less than [itex]\pi/2[/itex]. Since a is in the second quadrant, a= [itex]\pi[/itex]- .8208= 2.3208
 

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