Answer:
point of horizontal tangent is
and point of vertical tangent is
![(-15.17^(o),5.79)](https://img.qammunity.org/2020/formulas/mathematics/college/8xijxivyjdohgpycmxyfnj1vx9uh2jo132.png)
Explanation:
For a horizontal tangent it's slope should be zero thus
![r=6cos(\theta )\\\\(dr)/(d\theta)=-6sin(\theta )](https://img.qammunity.org/2020/formulas/mathematics/college/a9r3szungonn98zpts3eyilpponi4tlwtc.png)
Thus the ordered pair of
becomes (0,6) at this point tangent is horizontal
For a vertical tangent it's slope should be
Again differentiating the given curve we get
![r=6cos(\theta )\\\\ (dr)/(d\theta)=-6sin(\theta )](https://img.qammunity.org/2020/formulas/mathematics/college/byrk2573vqxz33omfb2h6nxdige79zt8lp.png)
![\therefore \theta =-15.17^(o)](https://img.qammunity.org/2020/formulas/mathematics/college/20w7pm1e322me3tjm555cr7zahk3ktdy81.png)
Thus the ordered pair of vertical tangent becomes (
![\theta =-15.17^(o),5.79)](https://img.qammunity.org/2020/formulas/mathematics/college/xvkun14a4hewv897awaspbuk8mc0lau7kf.png)