Show that the curves $r^2 = a^2\cos^2(\theta)$ and $r = a(1+\cos(\theta))$ intersect at an angle $3\arcsin((3/4)^{1/4})$.
I know how to solve this question but I am not getting the angle to be the given angle required in the question. Please help.
I tried getting the angle between the radius vector and the tangent vector for both the curves at the point of intersection and subtracting them, but I'm not getting the specified answer.
Indeed, your doubts are justified.
Let us consider the figure below. There are 2 intersection point $I$ and $J$. The figure being symmetrical with respect to the horizontal axis, let us work on point $I$.
The angle in point $I$ between the curves is by definition the angle between their tangents.
It has the plain value
$$\alpha = \pi/6+\pi/6=\pi/3 \ \ \tag{1}$$
as can be understood from the figure below:
Fig. 1: Common radius in green. Please note that here $a=1$ WLOG.
Indeed, the first equation, which can be written $r= \pm a\cos(\theta)$, gives a double circle, but in fact only its left circle (blue) with polar equation $r=-a \cos(\theta)$ needs to be considered.
The second equation corresponds to a cardioid (red curve).
The intersection point $I$ is easily found to be with polar coordinates
$$\begin{cases}r&=&1/2, & \theta&=&\theta_1&=&-\pi/3 & \text{for the circle}\\ r&=&1/2, & \theta&=&\theta_2&=&2\pi/3&\text{for the cardioid}\end{cases}$$
corresponding geometricaly to the third vertex of equilateral triangle $OIC$. With cartesian coordiantes $I(-1/4, \sqrt{3}/4)$ ; its symmetrical point being $J(-1/4,-\sqrt{3}/4)$.
For point $I$, the polar angles are $\theta_1=2\pi/3$ for the cardioid and $\theta_2=-\pi/3$ for the circle.
Knowing by differentiation that the leftmost point of the cardioid is precisely $I$ given that the tangent to the circle has polar angle $\theta_1=-\pi/3$by $\theta_2=2 \pi/3$ (yielding a vertical tangent), we get the result (1).
Remark: Of course, an analytical treatment is possible. Knowing formula $\tan \psi = \dfrac{r(\theta)}{r'(\theta)}$ (https://proofwiki.org/wiki/Angle_of_Tangent_to_Radius_in_Polar_Coordinates) giving the angle between the radius (depicted in green) and the tangent(s), one can retrieve the two angular values with values $\pi/6$ for the following values $\theta_1=2 \pi/3$ for the cardioid and $\theta_2=-\pi/3$ for the circle.