I need help at number 10 and 12 ?

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1 Answer
A08
Jul 10, 2017

(10). In the figure let line #AB# intersect line #CD# at #E#.
Let line #PE# be perpendicular drawn from #P# to the line #AB#.

Given is #anglePEB=90^@#
#=>angle PEA=90^@#
Now #angle PEA=angleDEA+anglePED#
Given #anglePED=(2/5x)^@#
Also #angle DEA=angleCEB=(3/4x-2)^@#, vertically opposite angles.

From above we get

#(3/4x-2)^@+(2/5x)^@=90^@#
#=>(3/4x+2/5x)^@=90^@+2^@#
#=>((15+8)/(4xx5)x)^@=92^@#
#=>((23)/(20)x)^@=92^@#
#=>x=92xx20/23#
#=>x=80^@#

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