Answer

  1. 60º
    Solution:
    In $\triangle\text{ABC},$
    $\angle\text{BCA}+\angle\text{CAB}+\angle\text{ABC}=180^\circ$
    $\Rightarrow3\text{y}^\circ+\text{x}^\circ+5\text{y}^\circ=180^\circ$
    $\Rightarrow8\text{y}^\circ+\text{x}^\circ=180^\circ\dots(1)$
    Also, $5\text{y}^\circ=180^\circ-7\text{y}^\circ$
    $\Rightarrow12\text{y}^\circ=180^\circ$
    $\Rightarrow\text{y}^\circ=15^\circ$
    From (1), $\text{x}^\circ=180^\circ-8\text{y}^\circ$
    $\Rightarrow\text{x}^\circ=180^\circ-8\times15^\circ$
    $\Rightarrow\text{x}^\circ=60^\circ$

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