搜索
11.（连云港中考）如图，在6×6正方形网格中，△ABC的顶点A，B，C都在网格线上，且都是小正方形边的中点，则sinA = ________.
答案：
$\frac{4}{5}$
12.（常州中考）如图，在四边形ABCD中，∠A = ∠ABC = 90°，DB平分∠ADC.若AD = 1，CD = 3，则sin∠ABD = ________.
答案：
$\frac{\sqrt{6}}{6}$
13.如图，在△ABC中，∠A = 120°，AB = 4，AC = 2，求sinB的值.
答案：
作$CD\perp BA$的延长线于点$D$，$\therefore \angle ADC = 90^{\circ}$，$\because \angle BAC = 120^{\circ}$，$\therefore \angle DAC = 180^{\circ}-\angle BAC = 60^{\circ}$，$\therefore \angle ACD = 30^{\circ}$，$\because AC = 2$，$\therefore AD = 1$，$CD = \sqrt{3}$，$\therefore BD = 5$，在$Rt\triangle BCD$中，$BC = \sqrt{BD^{2}+CD^{2}} = 2\sqrt{7}$，$\therefore \sin B=\frac{CD}{BC}=\frac{\sqrt{3}}{2\sqrt{7}}=\frac{\sqrt{21}}{14}$
14.如图，菱形ABCD的边长为10 cm，DE⊥AB，sinA = $\frac{3}{5}$，求DE的长和菱形ABCD的面积.
答案：
$\because DE\perp AB$，$\therefore \angle AED = 90^{\circ}$，在$Rt\triangle AED$中，$\sin A=\frac{DE}{AD}$，即$\frac{3}{5}=\frac{DE}{10}$，解得$DE = 6\ cm$，$\therefore$菱形$ABCD$的面积为$10\times6 = 60(cm^{2})$
15.如图，网格中的每个小正方形的边长都是1，△ABC的每个顶点都在网格的交点处，求sinA的值.
答案：
作$AD\perp BC$于点$D$，$CE\perp AB$于点$E$，由勾股定理得$AB = AC = 2\sqrt{5}$，$BC = 2\sqrt{2}$，$\therefore \triangle ABC$为等腰三角形，$\therefore D$为$BC$的中点，即点$D$在格点上，由勾股定理得$AD = 3\sqrt{2}$，由$\triangle ABC$的面积可得$BC\cdot AD = AB\cdot CE$，$\therefore CE=\frac{2\sqrt{2}\times3\sqrt{2}}{2\sqrt{5}}=\frac{6\sqrt{5}}{5}$，$\therefore \sin A=\frac{CE}{AC}=\frac{\frac{6\sqrt{5}}{5}}{2\sqrt{5}}=\frac{3}{5}$
16.如图，AB是半圆的直径，点O为圆心，OA = 5，弦AC = 8，OD⊥AC，垂足为E，交⊙O于点D，连接BE.设∠BEC = α，求sinα的值.
答案：
连接$BC$，$\because AB$是半圆的直径，$\therefore \angle ACB = 90^{\circ}$，在$Rt\triangle ABC$中，$AC = 8$，$AB = 10$，$\therefore BC = \sqrt{AB^{2}-AC^{2}} = 6$，$\because OD\perp AC$，$\therefore AE = CE=\frac{1}{2}AC = 4$，在$Rt\triangle BCE$中，$BE = \sqrt{BC^{2}+CE^{2}} = 2\sqrt{13}$，$\therefore \sin\alpha=\frac{BC}{BE}=\frac{6}{2\sqrt{13}}=\frac{3\sqrt{13}}{13}$
查看更多完整答案，请扫码查看
