A vertical tower OP stands at the centre O of a square ABCD. Let 'h' and 'b' denote the length OP and AB respectively. suppose (angle APB=60' then the relationship between 'h and 'b' can be expressed as:(a)2b² =h² (b)2h²=b² (c)3b²=2h² (d)3h²=2b²
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the triangle APB is an equilateral triangle. Because APB is 60 deg. ABP and PAB are 60 deg each, because AP = BP and hence ABP and PAB are both equal.
Now AOP is a right angle triangle. AO = b / root(2)
OP = h AP = b, and is the hypotenuse
b² = h² + b²/2 => 2 h² = b²
Now AOP is a right angle triangle. AO = b / root(2)
OP = h AP = b, and is the hypotenuse
b² = h² + b²/2 => 2 h² = b²
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Answered by
1
Answer:
Given−
ABCDisasquarewhosecentralpointisO.
OP⊥BO&∠APB=60
o
.AB=b&OP=h.
Tofindout−
Therelationbetweenb&h.
Solution−
OisthecentralpointofthesquareABCD.
∴OB=
2
1
×diagonal=
2
2
b
=
2
b
.........(i)
NowOisthecentralpointofthesquareABCD.
∴PA=PBi.e∠ABP=∠BAP.........(ii)
Again∠APB=60
o
.
∴∠ABP+∠BAP=180
o
−60
o
(byanglesumpropertyoftriangles)
∴⟹∠ABP+∠BAP=120
o
∴∠ABP=∠BAP=60
o
.(fromii)
SoΔPABisanequilateralone.
∴PB=AB=b.....(iii)
ConsideringΔPOBwehave∠O=90
o
,OB=
2
b
(fromi)
andPB=b(fromiii)
∴ApplyingPythagorastheorem
OP
2
+OB
2
=PB
2
⟹h
2
+(
2
b
)
2
=b
2
⟹2h
2
=b
2
Ans−OptionB
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