Respuesta :

Hrishii

Answer:

[tex]\huge{\red{\angle ABC = \boxed{30}\degree}}[/tex]

Step-by-step explanation:

  • [tex]\sin \angle ABC =\frac{p}{30}......(1)[/tex] (Given)

  • In [tex] \triangle ABC,[/tex] sin ratio of [tex] \angle ABC[/tex] can be given as:

  • [tex]\sin \angle ABC =\frac{AC}{AB}[/tex]

  • [tex]\implies \sin \angle ABC =\frac{2p+10}{80}......(2)[/tex]

  • From equations (1) and (2), we find:

  • [tex] \frac{p}{30}=\frac{2p+10}{80}[/tex]

  • [tex]\implies 80(p)=30(2p+10)[/tex]

  • [tex]\implies 80p=60p+300[/tex]

  • [tex]\implies 80p-60p=300[/tex]

  • [tex]\implies 20p=300[/tex]

  • [tex]\implies p=\frac{300}{20}[/tex]

  • [tex]\implies p=15[/tex]

  • [tex]\implies \sin \angle ABC =\frac{15}{30}[/tex]

  • [tex]\implies \sin \angle ABC =\frac{1}{2}[/tex]

  • [tex]\implies \sin \angle ABC =\sin 30\degree\:\:\:\:(\because \sin 30\degree=\frac{1}{2})[/tex]

  • [tex]\implies \huge{\red{\angle ABC = \boxed{30}\degree}}[/tex]

Otras preguntas