Help in trignometry! (?)

fiitjee ftre examfiitjee ftre exam

Calculator is not allowed in exams. I could find a sin inverse relation but that can't be solved without calculator.

1 Answer

75°

Explanation:

drawndrawn
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