HELP PLEASE!!!!!!!!!

asked Jun 24, 2023 25.3k views

5 votes

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3 votes

Part 1

$\angle C=180^(\circ)-44^(\circ)-35^(\circ)=\boxed{101^(\circ)}$

Part 2

$(b)/(\sin B)=(a)/(\sin A)\\\\b=(a \sin B)/(\sin A)\\\\b=(15 \sin 35^(\circ))/(\sin 44^(\circ))\\\\b \approx \boxed{12.385}$

Part 3

$(c)/(\sin C)=(a)/(\sin A)\\\\c=(a \sin C)/(\sin A)\\\\c=(15 \sin 101^(\circ))/(\sin 44^(\circ))\\\\c \approx \boxed{21.197}$

Maxlk answered Jun 28, 2023

by Maxlk

4.7k points

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