A triangle ABC and pqr are similar triangle such that angle is 32 and angle Rs 65 then find Angle B
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It is given that ∆ABC ~ ∆PQR\xa0Hence,\xa0{tex}\\angle {\\text{A}}{/tex}\xa0=\xa0{tex}\\angle {\\text{P}}{/tex}\xa0[by cpct]{tex}\\angle {\\text{B}}{/tex}\xa0=\xa0{tex}\\angle {\\text{Q}}{/tex}\xa0[by cpct]{tex}\\angle {\\text{C}}{/tex}\xa0={tex}\\angle {\\text{R}}{/tex}\xa0[by cpct]In ∆ABC, the sum of interior angles must be 180°{tex}\\angle A + \\angle B + \\angle C={/tex}\xa0{tex}180^{\\circ}{/tex}{tex}32^{\\circ}{/tex}+\xa0{tex}\\angle {\\text{B}}{/tex}\xa0+\xa0{tex}65^{\\circ}{/tex}=\xa0{tex}180^{\\circ}{/tex}{tex}97^{\\circ}{/tex}+{tex}\\angle {\\text{B}}{/tex} =\xa0{tex}180^{\\circ}{/tex}{tex}\\angle {\\text{B}}{/tex}\xa0=\xa0{tex}180^{\\circ}{/tex}-\xa0{tex}97^{\\circ}{/tex}{tex}\\therefore \\angle {\\text{B}}{/tex}\xa0=\xa0{tex}83^{\\circ}{/tex}