In the adjacent figure if x:y= …
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Sia ? 5 years, 5 months ago
According to question, x : y = 11 : 19
Therefore, x = 11a and y = 19 a
Now, AD is parallel to BE
So
{tex}\angle{/tex}BDC = {tex}\angle{/tex}ABD
y = 32
19 a = 32
So a = {tex}\frac{32}{19}{/tex}
Thus x = {tex}11 \times \frac{32}{19}{/tex}
x = 18.52
Since {tex}\angle{/tex}DCE ={tex}\angle{/tex}ADC
So. {tex}\angle{/tex}DCE = x + 32 = 18.52 + 32 = 50.32
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