A tower is broken at a point P above the ground. The top of the tower makes an angle 60° with the ground at Q. From another point R on the opposite side of Q angle of elevation of point P is 30°. If QR = 180 m, then what is the total height (in metres) of the tower?

In $$\triangle$$ PRS,

=> $$tan(30^\circ)=\frac{PS}{RS}$$

=> $$\frac{1}{\sqrt3}=\frac{x}{d}$$

=> $$d=\sqrt3x$$ ------------(i)

Similarly, in $$\triangle$$ PQS,

=> $$tan(60^\circ)=\frac{PS}{SQ}$$

=> $$\sqrt3=\frac{x}{180-d}$$

=> $$180\sqrt3-3x=x$$     [Using equation (i)]

=> $$x+3x=4x=180\sqrt3$$

=> $$x=\frac{180\sqrt3}{4}=45\sqrt3$$ ------------(ii)

Again, in $$\triangle$$ PQS,

=> $$sin(60^\circ)=\frac{PS}{PQ}$$

=> $$\frac{\sqrt3}{2}=\frac{x}{y}$$

=> $$\sqrt3y=2(45\sqrt3)$$       [Using equation (ii)]

=> $$y=\frac{90\sqrt3}{\sqrt3}=90$$ -----------(iii)

Adding equations (ii) and (iii), we get :

=> $$x+y=45\sqrt3+90$$

=> Height of tower = $$45(\sqrt3+2)$$ m

=> Ans - (D)