
解三角形ABC中,∠A=20度,∠B=80度,AB=DC,求∠ADB的度数?
2023年2月14日 · ∠bac=80=∠b,ac=bc, 以bc为边作正 bce,连ae, ∠abe=80-60=20=∠acd, ab=cd,be=bc=ac, abe≌ acd, ∠aeb=∠cad,∠ace=40,∠cae=∠cea=70, ∠aeb=10, ∠adb=20+10=30度。
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已知∠A≡20°,∠B=80°,AB=AC,AD=BC,求∠BCD的度数。 答案 …
答案是70度,要坐3条辅助线,求完整解答。 解:作∠BAE=∠B=80 (点E在直线AC的外侧),取AE=AB,连接DE、CE∵∠B=80, ∠A=20∴∠ACB=180-∠B-∠A=80∵.
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Bering Catamaran BC80 | Aluminium Expedition Long Range …
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在三角形ABC中,AB=AC,角BAC=80°O为三角形ABC内一点,且角…
2010年11月12日 · 在三角形abc中,ab=ac,角bac=80°o为三角形abc内一点,且角obc=10°,角oca=20°,求角bao的度数过a作ap⊥bc交co延长线于p,连pb。 由∠BAC=80°,AB=AC,∴∠ABC=∠ACB=50°,得:∠ABO=40°,∠OCB=30°,由AP⊥BC
In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to - Cuemath
In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to. Solution: From the question, In ∆ ABC, BC = AB and ∠B = 80°. As BC = AB, ∆ ABC is an isosceles triangle. Consider ∠C = ∠A = x. ∠B = 80° [Given] Using the angle sum property. We know that the sum of interior angles of a triangle = 180°. ∠A + ∠B + ∠C = 180°. Let us substitute the values.
如图,在 ABC中,∠BAC=80°,AB=AC,点P是ABC内一点, …
在bc下方取一点d,使得三角形abd为等边三角形,连接dp、dc,根据等边三角形的性质得到ad=ab=ac,求出∠dac、∠acd、∠adc的度数,根据三角形的内角和定理求出∠abc=∠acb=50°,即∠cdb=140°=∠bpc,再证 bdc≌ bpc,得到pc=dc,进一步得到等边 dpc,推出 apd≌ apc,根据 ...
如图,在 ABC中,AB=AC,∠BAC=80°,O为三角形内一 …
2022年2月22日 · 以bc为边,在a的同侧构建等边 bcd,连接ad, 因db=dc,ab=ac,ad=ad 所 以 dab≌ dac 因角bdc=60°,所以∠adb=30°, 因角bac=80°,ab=ac, 所以∠abc=50°, 所以∠dba=30° 在 dab与 ocb中 ∠dba=∠obc bd=bc ∠bda=∠bco=30° 所以 dab≌ ocb 所以bo=ba …
In triangle ABC, BC=AB and angle B= 80, then angle A=? - Brainly
In triangle ABC, BC=AB and angle B= 80, then angle A = 50. Solution: Given that in triangle ABC, BC = AB which means ABC is an isosceles triangle. We know the sum of angles in a triangle is 180°. Since it is an isosceles triangle the measure of two angles will be same. Let ∠A = x and ∠c = x. We know, Sum of all angles = 180°. x+x+80° = 180°.