The triangle ABC is such that AB=7cm, AC=5cm, BC=8cm and angle ABC=y.
Show that y = 38.2 degrees, correct to the nearest 0.1degree.
Calculate the area of triangle abc, giving your answer in cm^2 to three significant figures.
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You need to use the law of cosines to find if the angle y is of `38.2^o` . Notice that the angle y is the angle opposite the side AC, hence, using the law of cosines yields:
`AC^2 = AB^2 + BC^2 - 2AB*BC*cos y`
Substituting 5 for AC, 7 for AB and 8 for BC yields:
`25 = 49 + 64 - 112*cos y =gt 112 cos y = 88`
`cos y = 88/112 =gt cos y = 0.785`
Using an arrcos calculator yields that `y = 38.2793^o .`
You need to evaluate the area of triangle, hence you may use the following formula such that:
`A_(ABC) = (AB*BC*sin y)/2`
`A_(ABC) = (7*8*sin 38.2793)/2`
`A_(ABC) = 28*0.618`
`A_(ABC) = 17.304 cm^2`
Hence, using the law of cosines yields that the angle y = 38.2^o and evaluating the area of triangle yields `A_(ABC) = 17.304 cm^2.`
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