A statue 1.46m tall, stands on …

Let SP be the statue = 1.46 m(given)
Suppose PB be the pedestal = h metre
According to question angles of elevation of S and P are {tex}60 ^ { \circ }{/tex}and {tex}45 ^ { \circ }{/tex} respectively.
Further suppose AB = x m,

In right {tex}\Delta \mathrm { ABS },{/tex}
{tex}\frac { S B } { A B } = \tan 60 ^ { \circ } = \sqrt { 3 }{/tex}
{tex}\Rightarrow \frac { h + 1.46 } { x } = \sqrt { 3 }{/tex} ..............(i)
In right {tex}\Delta \mathrm { PAB },{/tex}
{tex}\frac { P B } { A B } = \tan 45 ^ { \circ } = 1{/tex}
{tex}\therefore{/tex} h = x ...................(ii)
Putting x = h in (i), we get
{tex}\frac { h + 1.46 } { h } = \sqrt { 3 } \Rightarrow h + 1.46 = \sqrt { 3 } h{/tex}
or h ({tex}\sqrt { 3 } - 1{/tex}) = 1.46 {tex}\therefore{/tex} {tex}h = \frac { 1.46 } { \sqrt { 3 } - 1 } \times \frac { \sqrt { 3 } + 1 } { \sqrt { 3 } + 1 }{/tex}
{tex}\therefore{/tex} {tex}h = \frac { 1.46 } { 2 } \times ( \sqrt { 3 } + 1 ) = 0.73 \times 2.732{/tex}
= 2m (nearly)
Thus, height of the pedestal = 2m