60 POINTSSSS Answer correctly please this depeneds on my dumb GRADE PLEASE
1 answer:
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Answer:
steps below
Step-by-step explanation:
3.2.1 AD = DB* sin 2 = DB * sin θ .. DE // AB ∠2= θ ... (1)
By laws of sines: DB / sin ∠5 = x / sin ∠4
∠4 = θ-α ∠5 = 180°-<u>∠1</u>-∠4 = 180°-<u>∠3</u>-∠4 = 180°-(90°-θ)-(θ-α)) = 90°+α
DB = (x*sin ∠5)/sin (θ-α)
= (x* sin (90°+α)) / sin (θ-α)
AD = DB*sinθ
= (x* sin (90°+α))*sinθ / sin (θ-α)
= x* (sin90°cosα+cos90°sinα)*sinθ / sin (θ-α) .... sin90°=1, cos90°=0
= x* cosα* sinθ / sin (θ-α)
3.2.2 Please apply Laws of sines to calculate the length
See the picture attached with letters to better understand the problem
in the right triangle ABD
applying the Pythagoras theorem
8²=AD²+(x+4)²-----> AD²=64-(x+4)²------> equation 1
in the right triangle ADC
applying the Pythagoras theorem
12²=AD²+(2x+1)²-----> AD²=144-(2x+1)²------> equation 2
equate equation 1 and equation 2
64-(x+4)²=144-(2x+1)²-------> 64-[x²+8x+16]=144-[4x²+4x+1]
64-x²-8x-16=144-4x²-4x-1
4x²-x²-8x+4x=144-1-64+16
3x²-4x=95
3x²-4x-95=0
using a graph tool----> to resolve the second order equation
see the attached figure N 2
the solution is
x=6.33
the answer is
x=6.33 units
in the right triangle ABD
applying the Pythagoras theorem
8²=AD²+(x+4)²-----> AD²=64-(x+4)²------> equation 1
in the right triangle ADC
applying the Pythagoras theorem
12²=AD²+(2x+1)²-----> AD²=144-(2x+1)²------> equation 2
equate equation 1 and equation 2
64-(x+4)²=144-(2x+1)²-------> 64-[x²+8x+16]=144-[4x²+4x+1]
64-x²-8x-16=144-4x²-4x-1
4x²-x²-8x+4x=144-1-64+16
3x²-4x=95
3x²-4x-95=0
using a graph tool----> to resolve the second order equation
see the attached figure N 2
the solution is
x=6.33
the answer is
x=6.33 units