Answer:
i. <DCB =
ii. Sin of <DCB = 0.8
Step-by-step explanation:
Let <DCB be represented by θ, so that;
Sin θ =
Thus from the given diagram, we have;
Sin θ =
= 0.8
This implies that,
θ = 0.8
= 53.1301
θ =
Therefore, <DCB = .
So that,
Sin of <DCB = Sin
= 0.8
Sin of <DCB = 0.8
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Work Shown:
Let x = HC and y = BH
AH = 12-x since AH = 12-HC and x = HC
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See the attached drawing below. The altitude BH forms three similar triangles which makes the proportion below possible
HC/BH = BH/AH
x/y = y/(12-x) ... substitution
x(12-x) = y^2 ... cross multiply
y^2 = -x^2+12x
note how I isolated y^2 instead of y. We will use this equation later.
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Focus on triangle AHB. Use the pythagorean theorem
(BH)^2 + (AH)^2 = (AB)^2
(y)^2 + (12-x)^2 = (9)^2
y^2 + 144-24x+x^2 = 81
-x^2+12x + 144-24x+x^2 = 81 ... replace y^2 with -x^2+12x
-x^2+12x + 144-24x+x^2-81 = 0
-12x + 63 = 0
-12x = -63
x = -63/(-12)
x = 5.25 <<-- answer in decimal form
x = 5 + 0.25
x = 5 + 1/4
x = 20/4 + 1/4
x = 21/4 <<-- answer as a fraction
