the length of the shadow of a verticle building on the ground increases by 40root3 metres when the inclination of the sun changes from 60 degrees to 45 degrees . the height of the building is?
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Ans: 40/(sqrt{3} - 1)
Case 1
Let height of building be D
Length of shadow be X
Tan60 = sqrt(3)
∴ D/X = sqrt(3)
D=Xsqrt(3)
Case 2
When ∠45
Tan 45 =1
D/(X+40sqrt(3))=1
From above D = Xsqrt(3)
Xsqrt(3)= X+40sqrt(3)
Hence, X= 40sqrt(3)/(sqrt (3)-1)
D=Xsqrt(3)=120/(sqrt(3)-1)
Case 1
Let height of building be D
Length of shadow be X
Tan60 = sqrt(3)
∴ D/X = sqrt(3)
D=Xsqrt(3)
Case 2
When ∠45
Tan 45 =1
D/(X+40sqrt(3))=1
From above D = Xsqrt(3)
Xsqrt(3)= X+40sqrt(3)
Hence, X= 40sqrt(3)/(sqrt (3)-1)
D=Xsqrt(3)=120/(sqrt(3)-1)
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doraksai:
thank u
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