Question

Find x,y,z

Answer 1

$$ {\textbf{Step -1: First consider the }}\mathbf{\Delta ABD }$$

$$ {\text{From given Trapezium }}\angle A + \angle C = {180^ \circ } $$

$$ {\text{and }}\angle B + \angle D = {180^ \circ } $$

$$ \text {for }\Delta ABD,\angle A + \angle B + \angle D = {180^ \circ } $$ $$\textbf{[sum of angles of a triangle is } \mathbf{180^ \circ ]}$$

$$ \Rightarrow {80^ \circ } + x + y = {180^ \circ } $$

$$ \Rightarrow x + y = {100^ \circ }{\text{ - - - - - - - (1)}} $$

$$ {\text{From Trapezium }}\angle A + \angle C = {180^ \circ } $$

$$ \Rightarrow {80^ \circ } + z = {180^ \circ } $$

$$ \Rightarrow z = {100^ \circ }{\text{ - - - - - - - - - - (2)}} $$

$$ {\textbf{Step -2: Now Consider }}\mathbf{\Delta BDC }$$

$$ For \Delta BDC $$

$$ \Rightarrow {30^ \circ } + z + (y - {30^ \circ }) = {180^ \circ } $$

$$ \Rightarrow {30^ \circ } + {100^ \circ } + y - {30^ \circ } = {180^ \circ } $$

$$ \Rightarrow y = {180^ \circ } - {100^ \circ } $$

$$ \Rightarrow y = {80^ \circ }{\text{ - - - - - - - - - - (3)}} $$

$$ {\text{Put the value of y in equation (1) , We get}} $$

$$ x + y = {100^ \circ } $$

$$ \Rightarrow x + {80^ \circ } = {100^ \circ } $$

$$ \Rightarrow x = {20^ \circ }{\text{ - - - - - - - - - - (4)}} $$

$$ {\textbf{Hence, }} \mathbf {x = {20^ \circ },y = {80^ \circ }{\textbf{ and }}z = {100^ \circ } }$$