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#
