Question

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Tứ giác ABCD có góc A = 60^o ; góc B = 90^o. Tính góc C, góc D nếu:

a, góc C - góc D = 20^o            b, góc C = \(\dfrac{3}{4}\) góc D

Answer 1

Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(\widehat{C}+\widehat{D}+60^0+90^0=360^0\)

=>\(\widehat{C}+\widehat{D}=210^0\)

a: ta có: \(\widehat{C}+\widehat{D}=210^0\)

\(\widehat{C}-\widehat{D}=20^0\)

Do đó: \(\widehat{C}=\dfrac{210^0+20^0}{2}=115^0;\widehat{D}=115^0-20^0=95^0\)

b: Ta có: \(\widehat{C}+\widehat{D}=210^0\)

=>\(\dfrac{3}{4}\cdot\widehat{D}+\widehat{D}=210^0\)

=>\(\dfrac{7}{4}\cdot\widehat{D}=210^0\)

=>\(\widehat{D}=210^0:\dfrac{7}{4}=120^0\)

\(\widehat{C}=\dfrac{3}{4}\cdot\widehat{D}=\dfrac{3}{4}\cdot120^0=90^0\)