Answer:
In figure (i), \[\angle A+\angle B+\angle C=180{}^\circ \] [since, sum of all angles of a triangle is 180°] \[90{}^\circ +a+70{}^\circ =180{}^\circ \] \[a+160{}^\circ =180{}^\circ \] \[a=180{}^\circ 160{}^\circ =20{}^\circ \] Since, c is an the exterior angle of ΔABD. ∴ \[\angle c=a+30{}^\circ =20{}^\circ +30{}^\circ =50{}^\circ \] \[[\because \angle exterior\text{ }angle\text{ }=\text{ }sum\text{ }of\text{ }interior\text{ }angles]\] Since, b is an the exterior angle of ΔADC. \[\therefore \angle b=60{}^\circ +70{}^\circ =130{}^\circ \] \[[\because \angle exterior\text{ }angles\text{ }=\text{ }sum\text{ }of\text{ }opposite\] \[\text{ }interior\,\,angles]\] In figure (ii), In \[\Delta PQS,\angle QPS+\angle PQS+\angle PSQ=180{}^\circ \] [since, sum of all angles of a triangles is 180°] \[55{}^\circ +60{}^\circ +a=180{}^\circ \Rightarrow 115{}^\circ +a=180{}^\circ \] ∴ \[a=180{}^\circ 115{}^\circ =65{}^\circ \] Now, a + b = 180° [since, linear pair has sum of 180°] \[65{}^\circ +\text{ }b\text{ }=180{}^\circ \] \[b=180{}^\circ 65{}^\circ =115{}^\circ \] In ΔPSR, \[\angle PSR+\angle SPR+\angle PRS=180{}^\circ \] [since, sum of all angles of a triangle is 180°] \[115{}^\circ +c+40{}^\circ =180{}^\circ \] \[c=180{}^\circ 155{}^\circ =25{}^\circ \]
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