Step-by-step explanation:
triangle ADA' and OPA' are similar so, using property of similar triangles :
6/AD = (13.3+3.6-PB'+2.1)/2.1
{A'P = 13.3 +3.6 - PB' + 2.1}
6/AD = (16.9-PB'+2.1)/2.1
6/AD = (19-PB')/2.1 - EQN I
As we know, AD = BC as both of them are Sam's height (constant)
triangle OPB' and BCB' are similar so, using property of similar triangles:
6/BC = PB'/3.6
OR, 6/AD = PB'/3.6 (AD = BC) - EQN II
Now,
comparing EQN I AND II we get :
(19-PB')/2.1 = PB'/3.6
or, 68.4 - 3.6PB' = 2.1PB'
or, 68.4 = 5.7PB'
so, PB' = 12 m
Now, as we know,
6/BC = PB'/3.6
OR, 6/BC = 12/3.6
OR, (6×3.6)/12 = BC
<h3>
so, BC = 1.8 m</h3><h3>
so, BC = 1.8 mso, AD = 1.8 m</h3>
So, the height of Sam is 1.8 m.
tanA' = AD/AA'
<A' = tan‐¹(1.8/2.1)
<B' = tan‐¹(1.8/3.6)
Now,
<A'OB' = 180 - tan‐¹(1.8/2.1) - tan‐¹(1.8/3.6)
<A'OB' = 112.833
<h3>
so, <A'OB' = 113° (approx.)</h3>
