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3 years ago

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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)

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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$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

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$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

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