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Zorluk: OrtaLaw of Sines and Law of Cosines

A landscape architect is designing a triangular courtyard garden. Two adjacent edges of the garden measure 88 meters and 1515 meters, and the angle between these two edges is 60∘60^\circ. What is the length, in meters, of the third edge of the garden?

Cevap: 13 meters

Cevap

The length of the third edge of the garden is 13 meters.
Applying the Law of Cosines c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C) with a=8a = 8, b=15b = 15, and C=60∘C = 60^\circ yields c2=82+152−2(8)(15)cos⁡(60∘)=64+225−240(0.5)=169c^2 = 8^2 + 15^2 - 2(8)(15)\cos(60^\circ) = 64 + 225 - 240(0.5) = 169. Taking the square root gives c=13c = 13 meters.

Adım Adım Çözüm

1
Identify the given side lengths and included angle.
Two sides are a=8 ma = 8\text{ m} and b=15 mb = 15\text{ m}, and their included angle is C=60∘C = 60^\circ.
The Law of Cosines is used when two sides and the included angle (SAS) are known.
2
Substitute the values into the Law of Cosines formula c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C).
c2=82+152−2(8)(15)cos⁡(60∘)=64+225−240(0.5)=169c^2 = 8^2 + 15^2 - 2(8)(15)\cos(60^\circ) = 64 + 225 - 240(0.5) = 169.
Evaluating the squared terms and trigonometric value cos⁡(60∘)=0.5\cos(60^\circ) = 0.5 simplifies the equation to c2=169c^2 = 169.
3
Solve for the side length cc by taking the square root.
c=169=13 metersc = \sqrt{169} = 13\text{ meters}.
The physical side length of a geometric figure must be positive.

Anahtar Kavram

Law of Cosines (c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C))
Tahmini Süre:1m 30s
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Konular

Law of Sines and Law of Cosines

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