Mathematics

In the adjoining Figure, $$\overline{XY}\parallel \overline {BC}$$. Find the values of $$x, y, z$$.


SOLUTION
Given that XY$$||$$BC
then,
In $$\Delta ABC$$
$$\angle A+\angle B+\angle C=180^o$$(property of triangle)
$$40^o+30^o+\angle C=180^o$$
$$70^o+z=180^o$$
$$z=180^o-70^o$$
$$z=110^o$$
$$XY||BC$$
then
$$\angle x^o=\angle B$$
$$\angle x^o=30^o$$.
Now, In $$\Delta AXY$$
$$\angle x^o+\angle y^o+\angle A=180^o$$(property of triangle)
$$30^o+\angle y^o=40^o=180^o$$
$$\angle y^o=180^o-70^o$$
$$\angle y^o=110^o$$
Hence, this is the answer.
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