Question

Given that tan(0) = -7/24 and theta is in Quadrant II, what is cos(theta)?

Answer 1

Given that

`\tan \theta=-(7)/(24)`

Where

`\tan \theta=(y)/(x)`

In the second quadrant x < 0, y > 0

`x=-24,y=7`

Where

`r^2=x^2+y^2^{}_{}`

Substitute for x and y to find the value of r

`\begin{gathered} r^2=(-24)^2+7^2=576+49=625^{} \\ r^2=625 \\ \text{square root of both sides} \\ \sqrt[]{r^2}=\sqrt[]{625} \\ r=25\text{ } \\ \text{With r}>0 \end{gathered}`

Since, it lies in the second quadrant,

`\cos \theta=(x)/(r)`

Substitute the values of x and r

`\cos \theta=-(24)/(25)`

Hence, the answer is

`\cos \theta=-(24)/(25)`