2025年通城学典通城1典中考复习方略数学江苏专用第78页答案
如图,在Rt△ABC中,∠C = 90°,BC = a,AC = b,AB = c
∠A的正弦:sin A = $\frac{∠A的对边}{斜边}$ = $\frac{a}{c}$
∠A的余弦:cos A = $\frac{∠A的邻边}{斜边}$ = ①____
∠A的正切:tan A = $\frac{∠A的对边}{∠A的邻边}$ = ②____

答案

①$\frac{b}{c}$;②$\frac{a}{b}$
在Rt△ABC中,∠C = 90°,BC = a,AC = b,AB = c,则
(1)三边关系:a² + b² = ③____
(2)两锐角关系:∠A + ∠B = ④____
(3)边角关系:sin A = cos B = $\frac{a}{c}$,cos A = sin B = $\frac{b}{c}$,tan A = $\frac{a}{b}$,tan B = $\frac{b}{a}$
(4)sin²A + cos²A = 1

答案

③$c^{2}$;④$90^{\circ}$
|已知条件|图形|解法|
|----|----|----|
|已知一直角边和一锐角(a,∠A)|![img id=5]|∠B = 90° - ∠A,c = $\frac{a}{sin A}$,b = $\frac{a}{tan A}$(或b = $\sqrt{c^{2}-a^{2}}$)|
|已知斜边和一锐角(c,∠A)|![img id=6]|∠B = 90° - ∠A,a = c · sin A,b = c · cos A(或b = $\sqrt{c^{2}-a^{2}}$)|
|已知两直角边(a,b)|![img id=7]|c = $\sqrt{a^{2}+b^{2}}$,由tan A = $\frac{a}{b}$求∠A,∠B = 90° - ∠A(或由tan B = $\frac{b}{a}$求∠B)|
|已知斜边和一直角边(c,a)|![img id=8]|b = $\sqrt{c^{2}-a^{2}}$,由sin A = $\frac{a}{c}$求∠A,∠B = 90° - ∠A(或由cos B = $\frac{a}{c}$求∠B)|
求B90A或由cosBfracacCmodelsa

答案