Question Number 193691 by cortano12 last updated on 18/Jun/23
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Commented by cortano12 last updated on 18/Jun/23
Answered by MM42 last updated on 18/Jun/23
$$\angle{PAQ}=\mathrm{28}\:\:\&\:\angle{AQP}=\mathrm{120} \\ $$$$\blacktriangle{APQ}\::\:\frac{{AP}}{{sin}\mathrm{120}}=\frac{\mathrm{10}}{{sin}\mathrm{28}}\Rightarrow{AP}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{{sin}\mathrm{28}} \\ $$$$\blacktriangle{ABP}\::\:{AB}={APsin}\mathrm{56}=\frac{\mathrm{5}\sqrt{\mathrm{3}}{sin}\mathrm{56}}{{sin}\mathrm{28}}=\mathrm{10}\sqrt{\mathrm{3}}{cos}\mathrm{28} \\ $$