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blagie [28]
3 years ago
5

ACD is a triangle and B is a point on AC. AB = 8cm and BC is 6cm. Angle BCD = 48° and angle BDC = 50°. (a) Find the length of BD

. (b) Find the length of AD. (c) Find the area of triangle ABD. (This is all one question)​

Mathematics
1 answer:
FromTheMoon [43]3 years ago
5 0

Answer:

  • 5.8206 cm
  • 10.528 cm
  • 23.056 cm^2

Step-by-step explanation:

(a) The Law of Sines can be used to find BD.

  BD/sin(48°) = BD/sin(50°)

  BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm

__

(b) We can use the Law of Cosines to find AD.

  AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°

  AD^2 ≈ 110.841

  AD ≈ √110.841 ≈ 10.5281 . . . cm

__

(c) The area of ∆ABD can be found using the formula ...

  A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°

  A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2

_____

Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.

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Step-by-step explanation:

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To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

f(r,h)=\frac{1}{3}\pi r^2 h

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We know for maximum volume r\neq 0. So let \lambda be the Lagranges multipliers be such that,

f_r=\lambda g_r

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And,

f_h=\lambda g_h

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Substitute (3) in (2) we get,

\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})

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\pi r\sqrt{h^2+r^2}=8

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V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114

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