Let the curves be # C_1 : y=5x^3, and, C_2 : y=5x^2#.

To find their pt. of intersection, we have to solve their eqns.

#5x^3=y=5x^2 rArr 5x^2(x-1)=0 rArr x=0, x=1#.

#y=5x^2, x=0, x=1 rArr y=0, y=5#. Thus, we have,

#C_1nnC_2={O(0,0), A(1,5)}#.

To find the slopes of tgts. to the curves #C_1, and, C_2#, at the pts. #O, and, A#, we will find #dy/dx# as it denotes the reqd. slopes.

For #C_1, dy/dx=15x^2, &, "for" C_2, dy/dx=10x#

#:. for C_1, and, C_2, [dy/dx]_O=0.#

This means that, both tgts. to #C_1,&,C_2# have the same slope, namely #0#, and, both pass thro. the same pt. #O(0,0)#.

Clearly, **(1)** the #X#-axis is the **common tgt.** to #C_1 and C_2# touching at the Origin. **(2)** The angle btwn. them, at #O# is #0^@#.

# "At the pt." A(1,5), dy/dx=15, "for" C_1, and, "for" C_2, dy/dx=10#

:. #m_1=15, and m_2=10,# are the respective slopes of tgts.

#:. alpha=the /_(C_1,C_2) rArr tan alpha=|(m_1-m_2)/(1+m_1m_2)|#

#=|(15-10)/(1+150)|=5/151~~0.0331#

#:. alpha=arc tan 0.0331=1.8965^@#.

Enjoy Maths.!