解三角形高考(三道关于解三角形的高一数学问题)

1.

a=4，b=3，C=60°

c^2=a^2+b^2-2abcosC=4^2+3^2-2*4*3*1/2 = 13

c=√13

sinA=asinC/c = 4*√3/2 /√13 = 2√39 /13

2.

b^2=ac

b^2=a^2+c^2-2accosB

ac=a^2+c^2-2accosB

cosB = (a^2+c^2-ac)/(2ac) = (a^2+c^2)/(2ac) - 1/2

∵(a-c)^2≥0，a^2+c^2≥2ac，(a^2+c^2)/(2ac) ≥ 1

∴cosB = (a^2+c^2)/(2ac) - 1/2 ≥ 1/2

∴0＜B ≤ 60°

3.

(b+c):(c+a):(a+b)=12:8:10

(b+c)/(c+a)=12/8，(b/c+1)/(1+a/c)=3/2，2b/c-3a/c=1 ... (1)

(c+a)/(a+b)=8/10，(1+a/c)/(a/c+b/c)=4/5，4b/c-a/c=5 ...(2）

(2)-(1)*2得：5a/c=3，a/c = 3/5

4b/c=5+a/c=5+3/5=28/5，b/c=7/5

cosB = (a^2+c^2-b^2)/(2ac)

= { (a/c)^2 + 1 - (b/c)^2 } / (2a/c)

= { (3/5)^2 + 1 - (7/5)^2 } / (2*3/5)

= ( 3^2 + 5^2 - 7^2 ) / (2*3*5)

= -15/30

= -1/2

B=120°