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2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

                

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Given AB || CD and CD || EF and y:z = 3:7 \Rightarrow z = \frac{7}{3}y
therefore, AB || EF and x =z (alternate interior angles)..............(i)

Again, CD || AB 
\\\Rightarrow x+y =180^0\\ \Rightarrow z+y =180^0.............(ii)

Put the value of z in equation (ii), we get

\frac{10}{3}y =180^0 \Rightarrow y =54^0
Then z=180^0-54^0=126^0

By equation (i), we get the value of x=126^0
 

Posted by

manish painkra

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Posted by

Kantilal Thakur

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