Answer:
[tex]\csc(\theta)=\sqrt{2}[/tex]
Step-by-step explanation:
Please refer to the attachment below.
We have:
[tex]5\cos^2(\theta)+6\sin^2(\theta)=11/2[/tex]
Remember that cosine is the ratio of the adjacent side to the hypotenuse. Therefore:
[tex]\cos(\theta)=\frac{b}{c}[/tex]
And sine is the ratio of the opposite side to the hypotenuse. Therefore:
[tex]\sin(\theta)=\frac{a}{c}[/tex]
Substitute into our original equation:
[tex]5(\frac{b}{c})^2+6(\frac{a}{c})^2=11/2[/tex]
Square:
[tex]\frac{5b^2}{c^2}+\frac{6a^2}{c^2}=\frac{11}{2}[/tex]
Multiply both sides by c squared:
[tex]5b^2+6a^2=\frac{11}{2}c^2[/tex]
Let’s also multiply both sides by 2 to eliminate the fraction:
[tex]10b^2+12a^2=11c^2[/tex]
Remember that according to the Pythagorean Theorem:
[tex]c^2=a^2+b^2[/tex]
Hence, we can make another substitution:
[tex]10b^2+12a^2=11(a^2+b^2)[/tex]
Distribute:
[tex]10b^2+12a^2=11a^2+11b^2[/tex]
Subtract 10b² from both sides:
[tex]12a^2=11a^2+b^2[/tex]
Subtract 11a² from both sides:
[tex]a^2=b^2[/tex]
Take the square root of both sides:
[tex]a=b[/tex]
Hence, a is equivalent to b.
Then, we must have a 45-45-90 Triangle.
Therefore, our angle θ is 45°.
We can substitute this back into our original equation to make sure. Remember that cos(45)=sin(45)=√2/2:
[tex]\begin{aligned} 5\cos^2(45)+6\sin^2(45)&=11/2 \\ 5(\frac{\sqrt{2}}{2}})^2+6(\frac{\sqrt{2}}{2})^2&=11/2 \\ 5(2/4)+6(2/4)&=11/2 \\ 5(1/2)+6(1/2)&=11/2 \\ 5/2+6/2&=11/2 \\ 11/2&=11/2 \;\checkmark \end{aligned} \\[/tex]
Hence:
[tex]\begin{aligned} \sin(45)&=\frac{1}{\sqrt{2}} \\ \Rightarrow \csc(\theta)&=\sqrt{2} \end{aligned}[/tex]
