Yahoo Answers: Answers and Comments for I need some help on this geometry problem, its really confusing. Please and thank you? [Homework Help]
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enUS
Sat, 19 Jun 2010 16:44:08 +0000
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Yahoo Answers: Answers and Comments for I need some help on this geometry problem, its really confusing. Please and thank you? [Homework Help]
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https://answers.yahoo.com/question/index?qid=20100619164408AAltb1n
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From Horatio: In spite of what you posted, the answer select...
https://answers.yahoo.com/question/index?qid=20100619164408AAltb1n
https://answers.yahoo.com/question/index?qid=20100619164408AAltb1n
Wed, 23 Jun 2010 17:03:31 +0000
In spite of what you posted, the answer selections DO match up with this problem.
Drawing the kite with those dimensions, you have a deltoid quadrilateral. Make sure you draw the 4 foot dimension is the height of the kite, from corner to top corner.
Therefore, this forms two congruent triangles that have one short side 2 feet long, another side 3 feet long, and a large side 4 feet long. The large side is shared between the two congruent triangles.
Look at the figure in the source link I posted below. The 2 foot sides are AB and AD. The 3 foot sides are BC and DC. The 4 foot side is AC.
The two congruent triangles are ABCA and ACDA.
Since we know the three sides of both triangles, we use the Law of Cosines to calculate angles.
The Law of Cosines state C² = A² + B²  2AB*cosθ where θ is the angle opposite side C. In the figure at the source link I posted below, this angle is angle ABC and angle ADC.
In this problem, if you assign 2 feet to A and 3 feet to B and 4 feet to C, then θ will be the angle opposite the 4 foot side. That means that θ is the angle formed between the 2 foot and 3 foot sides, and that is what this problem is asking us to calculate.
solving for cosθ, we have
cosθ = (c²  a²  b²)/2ab
cosθ = (c² + a² + b²)/2ab
cosθ = (a² + b²  c²)/2ab
Taking the inverse cosine (arccos) of both sides of the equal sign, we have:
θ = arccos((a² + b²  c²)/2ab) where θ is in degrees
So plugging in the values we already know;
θ = arccos((2² + 3²  4²)/(2*2*3))
θ = arccos((4 + 9  16)/(12))
θ = arccos(3/12)
θ = 104.5º (the angle between the 2 foot side and the 3 foot side)
Now compare this answer (104.5º) with the answer selections you posted above. Can you find it now?