题目 #19420

∵Rt △ABC 中，AC = 4，BC = 3，AB = 5， ∴sinB = _{AB}=\(AC _{5}\).^{4} 2.B． A.sinA = a b\(a _{c}\), 结论错误, 故不符合题意; B．tanB = _{a}, 结论正确,故符合题意; C．tanA = _{b}, 结论错误, 故不符合题意; D．cosB =\(a_{c}\),结论错误, 故不符合题意; 故选; B． 3.B 1 ∵2sin∠A = 1, ∴sin∠A = _{2}, ∠A = 30∘, 故选: B. 4.C. 35 ∵△ABC中，∠C = 90°，a, b, c分别是∠A, ∠B, ∠C对边，∵3a = 4b，∴令b = 3x, a = 4x, c = 5x, ∴sinB = _{c}= _{5x}= _{5}.^{3} b 3x 5.D 如图,过点B作BQ ⊥AD于点Q, 因为在菱形ABCD中, 对角线AC, BD相交于点O, AC = 8, BD = 6，∴ 1\(S_{ABCD}\)= _{2}AC ⋅BD = 24, OA = OC = 4, OB =OD=3, AB = AD, ∴AB =\(\sqrt{3^{2}}\)+\(4^{2}\)= 5 = AD, ∴AD ⋅BQ = 24, ∴ 24 BQ\(\frac{24}{5} 24 _{5}= _{25}\). 故选: D. BQ = _{5},所以sin∠BAD = _{AB}= 6.C ∵角的正弦值\(\sqrt{3_{2}}\), sin\(\alpha =√3_{2}\)∴锐角\(\alpha = 60∘\)故选C. 4 7. 5 因为在Rt △ABC中, ∠A = 90°, AB = 3, AC = 4, ∴BC =\(\sqrt{AB^{2}}\)+\(AC^{2}\)= 5,所以sinB =\(AC_{BC}\)= _{5}.^{4} 1 8. 4 如图取格点D, 连接AD, CD, 由网格的特征可知, ∠CAD = 45∘, ∠BAC = 90∘+45°= 135∘, ∴∠CAD + ∠BAC = 180∘, ∴D, A, B三点共线, 由格点三角形可知:AD =\(\sqrt{1^{2}}\)+\(1^{2}\)= √2, CD =\(\sqrt{1^{2}}\)+\(1^{2}\)= √2, AC = 2, ∴\(AD^{2}\)+\(CD^{2}\)=\(AC^{2}\),所以\(\Delta{}ACD\)是直角三角形, ∠ADC = 90∘, ∵CD = √2, BD =\(\sqrt{4^{2}}\)+\(4^{2}\)= 4√2,所以tan∠ABC =\(CD _{BD}\)=\(1 _{4}\), 故答案为: _{4}.^{1} 7 9. 9 过点 A 作 AD ⊥BC，垂足为 D，过点 C 作 CE ⊥AB，垂足为 E，在Rt △ADB 中，cos∠B = BD\(1 _{AB}\)= _{3}, ∴BD = 1\(1 _{3}\)AB = 1. ∵AB = AC, AD ⊥BC,所以BD = DC, ∴BC = 2, AD =\(\sqrt{AB^{2}}\)−\(BD^{2}\)=\(\sqrt{3^{2}}\)−\(1^{2}\)= 2√2,因为_{2}AB ⋅ 7 AE\(\frac{7}{3}\)CE = _{2}BC ⋅AD, ∴CE = AB = 3 = _{3},所以AE =\(\sqrt{AC^{2}}\)−\(CE^{2}\)= _{3}，∴cos∠BAC = _{AC}= _{3}= _{9}.^{7} 1 BC⋅AD\(2\times 2√2 4√2 2 10\). 3 因为D 为 AB 的中点, DE ∥AC, ∴AC = 2DE, ∵DE = 1\(1 _{3}\)BD, BD = AD,所以DE = _{3}AD, AD = 3DE, 在Rt\(\Delta{}ABC\)中 , cosA = _{AD}= _{3DE}= _{3}.^{2} AC 2DE 11.√22 如图所示, 连接BE, 由题可知此时BE ⊥AC, 在网格中, 由勾股定理可得:AE =\(\sqrt{1^{2}}\)+\(3^{2}\)= √10, AB =\(\sqrt{2^{2}}\)+\(4^{2}\)= 2√5, ∴在Rt\(\Delta{}ABE\)中，cos∠CAB= _{AB}=√10AE2√5=\(\sqrt{2_{2}}\). 故答案为:\(\sqrt{2_{2}}\)． 12.45∘ ∵\(sin^{2}\)A =\(1_{2}\), sinA > 0 ∴sinA =\(\sqrt{2_{2}}\), ∴∠A = 45∘, 故答案为45∘. 3 13. 4 ∵ 在 Rt △ABC 中，∠C = 90°, tanA =\(BC_{AC}\), BC = 6, AC = 8, ∴tanA = _{AC}= _{4}.^{3} BC 10√221 14. 221 由勾股定理得: AC =\(\sqrt{2^{2}}\)+\(3^{2}\)= √13, BC =\(\sqrt{1^{2}}\)+\(4^{2}\)= √17. 如图,作AD ⊥BC于点D, 则∠ADC = 90∘, ∵ 1 1 1 1\(2\times 5 10√17\)S△ABC=\(3 \times 4\)− _{2}\(\times 2 \times 2\)− _{2}\(\times 2 \times 3 = 5\)∴ _{2}BC ⋅AD = 5, ∴AD = √17= _{17},所以sin∠ACB = 10√17 10\(\sqrt{\frac{_{2}\times 1\times 4 −}{_{221}.}221}\)AD 17 10√221,故答案为: _{AC}= √13= 221 15.B 因为√2sin(\(\alpha + 20∘\)) = 1, 所以sin(\(\alpha + 20∘\)) =\(\sqrt{2_{2}}\), 所以锐角\(\alpha + 20∘= 45∘\)则\(\alpha = 45∘−20∘= 25∘\)故选B.