Question

Please help, screenshot is here:

Answer 1

`Answer:\ cos(\theta_1)=\boxed{-(40)/(41) }`

Explanation:

`\displaystyle\\sin(\theta_1)=(9)/(41)\ \ \ \ \ \ 90^0 < \theta_1 < 180^0\ \ \ \ \ \ \ cos(\theta_1)=?\\\\sin^2(\theta_1)+cos^2(\theta_1)=1\\\\cos^2(\theta_1)=1-sin^2(\theta_1)\\\\cos^2(\theta_1)=1-((9)/(41))^2\\\\cos^2(\theta_1)=1-(81)/(1681) \\\\cos^2(\theta_1)=(1(1681)-81)/(1681) \\\\cos^2(\theta_1)=(1600)/(1681) \\\\`

Extract the square root of both parts of the equation:

`\displaystyle\\\cos(\theta_1)=б\sqrt{(1600)/(1681) } \\\\\cos(\theta_1)=б\sqrt{(40^2)/(41^2) } \\\\\cos(\theta_1)=б\sqrt{((40)/(41))^2 } \\\\\cos(\theta_1)=б(40)/(41)`

`\displaystyle\\Since\ 90^0 < \theta_1 < 180^0, \\\\ therefore\ cos(\theta_1)=-(40)/(41)`