Question

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Answer 1

Bài 3:

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

=>\(\hat{A}+\hat{D}=360^0-46^0-80^0=234^0\)

=>\(2\cdot\hat{A}+\hat{A}=234^0\)

=>\(3\cdot\hat{A}=234^0\)

=>\(\hat{A}=\frac{234^0}{3}=78^0\)

\(\hat{D}=2\cdot\hat{A}=2\cdot78^0=156^0\)

b: Xét tứ giác MNPQ có \(\hat{M}+\hat{N}+\hat{P}+\hat{Q}=360^0\)

=>\(\hat{P}+\hat{Q}=360^0-113^0-59^0=188^0\)

=>\(\frac13\cdot\hat{Q}+\hat{Q}=188^0\)

=>\(\frac43\cdot\hat{Q}=188^0\)

=>\(\hat{Q}=188^0:\frac43=188^0\cdot\frac34=141^0\)

\(\hat{P}=141^0\cdot\frac13=47^0\)

Bài 1:

Hình 1:

\(\hat{NPQ}+60^0=180^0\)

=>\(\hat{NPQ}=180^0-60^0=120^0\)

Xét tứ giác MNPQ có \(\hat{M}+\hat{MNQ}+\hat{MQP}+\hat{NPQ}=360^0\)

=>\(\hat{M}=360^0-60^0-120^0-100^0=80^0\)

Hình 2:

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

=>\(\hat{B}+\hat{C}+\hat{D}=360^0-120^0=240^0\)

mà \(\hat{B}=\hat{C}=\hat{D}\)

nên \(\hat{B}=\hat{C}=\hat{D}=\frac{240^0}{3}=80^0\)

Bài 2:

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

=>\(\hat{A}+\hat{D}=360^0-56^0-124^0=180^0\)

mà \(\hat{A}-\hat{D}=30^0\)

nên \(\hat{A}=\frac{180^0+30^0}{2}=105^0;\hat{D}=105^0-30^0=75^0\)

b: Xét tứ giác MNPQ có \(\hat{M}+\hat{N}+\hat{\left.P\right.}+\hat{Q}=360^0\)

=>\(\hat{P}+\hat{Q}=360^0-68^0-122^0=170^0\)

mà \(\hat{P}-\hat{Q}=30^0\)

nên \(\hat{P}=\frac{170^0+30^0}{2}=100^0;\hat{Q}=100^0-30^0=70^0\)

c: Xét tứ giác EHGK có \(\hat{E}+\hat{H}+\hat{G}+\hat{K}=360^0\)

=>\(\hat{G}+\hat{K}=360^0-60^0-60^0=240^0\)

mà \(\hat{G}-\hat{K}=120^0\)

nên \(\hat{G}=\frac{240^0+120^0}{2}=180^0\) (vô lý)

=>Đề sai rồi bạn

d: \(\hat{B}=2\cdot\hat{A}=2\cdot50^0=100^0\)

Xét tứ giác ABEG có \(\hat{A}+\hat{B}+\hat{E}+\hat{G}=360^0\)

=>\(\hat{E}+\hat{G}=360^0-50^0-100^0=210^0\)

mà \(\hat{E}-\hat{G}=40^0\)

nên \(\hat{E}=\frac{210^0+40^0}{2}=\frac{250^0}{2}=125^0\)

Bài 4:

a:

Xét ΔKLI và ΔKJI có

KL=KJ

LI=JI

KI chung

Do đó: ΔKLI=ΔKJI

=>\(\hat{KLI}=\hat{KJI}\)

=>\(\hat{KJI}=100^0\)

Xét tứ giác KLIJ có \(\hat{K}+\hat{L}+\hat{I}+\hat{J}=360^0\)

=>\(\hat{K}=360^0-55^0-100^0-100^0=360^0-200^0-55^0=160^0-55^0=105^0\)

b: Đặt \(a=\hat{A};\hat{B}=b;\hat{C}=c;\hat{D}=d\)

\(\hat{A}:\hat{B}:\hat{C}:\hat{D}=2:5:4:1\)

=>a:b:c:d=2:5:4:1

=>\(\frac{a}{2}=\frac{b}{5}=\frac{c}{4}=\frac{d}{1}\)

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

=>\(a+b+c+d=360^0\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a}{2}=\frac{b}{5}=\frac{c}{4}=\frac{d}{1}=\frac{a+b+c+d}{2+5+4+1}=\frac{360^0}{12}=30^0\)

=>\(a=30^0\cdot2=60^0;b=30^0\cdot5=150^0;c=30^0\cdot4=120^0;d=30^0\cdot1=30^0\)

=>\(\hat{A}=60^0;\hat{B}=150^0;\hat{C}=120^0;\hat{D}=30^0\)