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1.三角形ABC之三邊長為a=BC=8,b=CA=7,c=AB=5,求:

(1)角B=?

使用餘弦定律:

B=acos{(a^2+c^2-b^2)/(2ac)}

=acos{(64+25-49)/(80)}

=acos{40/80}

=acos(1/2)

=60(deg)

(2)三角形ABC之面積為?

使用Δ面積公式: s=(a+b+c)/2=10

Δ=√{s(s-a)(s-b)(s-c)}

=√(10*2*3*5)

=10√3

(3)三角形ABC之外接圓半徑為?

使用正弦定律:

a/sinA=b/sinB=c/sinC=2R

R=b/(2*sinB)

=7/(2*sin60)

=7/√3

=7√3/3

(4)三角形ABC之內切圓半徑為?

使用內切圓公式:

r=Δ/s

=10√3/10

=√3

(5)BC邊上之中線長為?

使用座標法: B=(0,0), C=(8,0)

A=5(cosB,sinB)

=5(cos60,sin60)

=5(1/2,√3/2)

=(2.5,2.5√3)

D=BC中點

=(B+C)/2

={(0,0)+(8,0)}/2

=(4,0)

AD^2=(A-D)^2

=(2.5-4)^2+(2.5√3-0)^2

=1.5^2+3*2.5^2

=2.25+3*6.25

=21

AD=√21.....ans

2.三角形ABC中,cosA=4/5,cosB=2/√5,c=AB=√5,則三角形的外接圓半徑為?

(A)5/4 (B)5/2 (C)15/4 (D)5根號5/4 (E)9根號5/4

sinA=√{1-cos^2(A)}

=√{1-16/25}

=√(9/25)

=3/5

sinB=√{1-cos^2(B)}

=√{1-4/5}

=√(1/5)

=1/√5

使用正弦定律:

a/sinA=b/sinB=c/sinC=2R

b=a*sinB/sinA

=a*5/3√5

=a√5/3

投影定律:

c=a*cosB+b*cosA

√5=a*2/√5+b*4/5 => a=3/2, b=√5/2

R=a/(2*sinA)

=(3/4)*5/3

=5/4.....ans=(A)

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