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练练练案案案及及及考考考案案案部部部分分分
详详详解解解答答答案案案
[练案部分]
练案[1]
A 组 素养自测
1. C 由正弦定理 asin A =
b
sin B,得
5
1
2
= 15sin B,
∴ sin B = 32 .
∵ b > a,∴ B = 60°或 B = 120° . 当 B = 60°时,C = 90°,此时 c = 2 5 . 当 B =
120°时,C = 30°,此时 c = a = 5 . 故选 C.
2. B 由 asin A =
b
sin B,知
3
1
3
= 5sin B,即 sin B =
5
9 ,选 B.
3. A 由已知,得 32 =
1
2 × 2 × 3 × sin A,∴ sin A =
3
2 .
4. B 由已知,∠C = 30°,由正弦定理 csin C =
b
sin B,解得 c = 2.
5. BD 由正弦定理可知,B、D 正确.
6. 2 在△ABC 中,∠A = 75°,∠B = 45°,所以∠C = 60°,
由正弦定理知
AC
sin B =
AB
sin C,
所以 AC = ABsin Bsin C =
6 × sin 45°
sin 60° = 2.
7. 12 因为 A + B + C = 180°,且 A + C = 2B,所以 B = 60°,
由正弦定理得 sin A = asin Bb =
1 × sin 60°
3
= 12 .
8. 1 由 ac =
sin A
sin C,得 sin C =
1
2 ,所以 C = B =
π
6 ,
所以
b
c =
sin B
sin C = 1.
9. 由正弦定理得,sin C = csin Bb =
2sin 30°
2
= 22 ,
∵ c > b,0° < C < 180°,∴ C = 45°或 135°. 当 C = 45°时,A = 105°,
∴ a = bsin Asin B =
2sin 105°
sin 30° = 3 + 1.
当 C = 135°时,A = 15°,
∴ a = bsin Asin B =
2sin 15°
sin 30° = 3 - 1.
10. ∵ cos B2 =
2 5
5 ,∴ cos B = 2cos
2 B
2 - 1 =
3
5 .
∴ sin B = 45 .
∵ C = π4 ,
∴ sin A = sin(B + C) = sin Bcos C + cos Bsin C = 7 210 .
∵ asin A =
c
sin C,∴ c =
asin C
sin A =
2
7 2
10
× 22 =
10
7 .
∴ S = 12 acsin B =
1
2 × 2 ×
10
7 ×
4
5 =
8
7 .
B 组 素养提升
1. B 根据正弦定理,得sin Aa =
sin B
b . 由已知得
sin B
b =
cos A
a ,∴
sin A
a =
cos A
a ,
∴ sin A = cos A,∴ tan A = 1.
又 0 < A < π,∴ A = π4 .
2. D 由正弦定理,得 c = asin Csin A = 2,∵ B = 180° - 30° - 45° = 105°,
sin 105° = sin(60° + 45°) = sin 60°cos 45° + cos 60°sin 45° = 6 + 24 ,
∴ S△ABC =
1
2 acsin B =
3 + 1
4 .
3. AC 由正弦定理可知 asin A =
b
sin B,
∴ sin B = bsin Aa =
2 3 × 12
2 =
3
2 ,
∵ B∈(0°,180°),∴ B = 60°或 120°.
4. B 因为 a = 2,c = 2,
所以由正弦定理可知, 2sin A =
2
sin C,故 sin A = 2sin C,
又 B = π - (A + C),
故 sin B + sin A(sin C - cos C) = sin(A + C) + sin Asin C - sin Acos C
= sin Acos C + cos Asin C + sin Asin C - sin Acos C = (sin A + cos A)sin C = 0.
又 C 为△ABC 的内角,故 sin C≠0,则 sin A + cos A = 0,即 tan A = - 1.
又 A∈(0,π),所以 A = 3π4 .
从而 sin C = 1
2
sin A = 22 ×
2