The above figure explains the situation described in the given problem.

#Q->"the starting point"#

#P->"the point at a distance d m from from Q"#

#R->"the point from which angle of depression of P is " 30^@#

#S->"the point from which angle of depression of P is " 60^@#

Let the angle of inclination of slope of the mountain be #theta#

Here #QR=c and QS=2c#

So #RU=csintheta ;QU=c costheta#

and #ST=2csintheta ;QT=2c costheta#

Now #(RU)/(PU)=tan30^@#

#=>(RU)/(QU+QP)=tan30^@#

#=>(csintheta)/(c costheta+d)=1/sqrt3#

#=>c costheta+d=sqrt3csintheta......[1]#

Again #(ST)/(PT)=tan45^@#

#=>(ST)/(QT+QP)=tan45^@#

#=>(2csintheta)/(2c costheta+d)=1#

#=>2c costheta+d=2csintheta......[2]#

Subtracting (1) from (2) we get

#2csintheta-sqrt3csintheta=c costheta#

#=>csintheta(2-sqrt3)=c costheta#

#=>tantheta=1/(2-sqrt3)=2+sqrt3#

#=>theta =75^@#

**Alternative method**

In #Delta PQR,angleQRP=(theta-30^@)#

So #c/(sin30^@) =d/sin(theta-30^@)#

#=>2c=d/sin(theta-30^@)#

And in #Delta PQS,angleQSP=(theta-45^@)#

So #(2c)/(sin45^@) =d/sin(theta-30^@)#

#=>2sqrt2c=d/sin(theta-45^@)#

Hence #sqrt2sin(theta-45^@)=sin(theta-30^@)#

#=>sintheta-costheta=sqrt3/2sintheta-1/2costheta#

#=>2sintheta-2costheta=sqrt3sintheta-costheta#

#=>tantheta=1/(2-sqrt3)=2+sqrt3=tan75^@#

**Please note**

#tan75^@=tan(45^@+30^@)#

#=(tan45^@+tan30^@)/(1-tan45^@tan30^@)#

#=(1+1/sqrt3)/(1-1/sqrt3)#

#=(sqrt3+1)/(sqrt3-1)#

#=(sqrt3+1)^2/((sqrt3)^2-1^2)#

#=(4+2sqrt3)/2=2+sqrt3#

Describe your changes (optional) 200