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Question Number 203778 by mr W last updated on 27/Jan/24

Answered by mr W last updated on 27/Jan/24

Commented by mr W last updated on 27/Jan/24

$$anotherway$$$$AD//EB$$$$\frac{AD}{EB}=\frac{3}{3+5}$$$$\Rightarrow AD=\frac{3}{8}\times 5=\frac{15}{8}$$

Answered by esmaeil last updated on 27/Jan/24

$$\frac{1}{2}AC\times ADsin60+\frac{1}{2}AB\times ADsin60=$$$$\frac{1}{2}AB\times ACsin120\to$$$$3AD+5AD=15\to AD=\frac{15}{8}$$

Commented by mr W last updated on 27/Jan/24

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Commented by esmaeil last updated on 27/Jan/24

$$thanks.$$$$youarewonderfulilearnedalot$$$$fromyou.$$

Answered by ajfour last updated on 28/Jan/24

Commented by ajfour last updated on 28/Jan/24

$$here\frac{h}{k}=\frac{5}{3},b=5,c=3$$$$a^2=b^2+c^2-2bc\cos \frac{2\pi }{3}=25+9+15$$$$hence$$$$2s^2=34-\frac{16}{4}-\frac{30}{64}(25+9+15)$$$$=30-\frac{15}{32}\times 49=\frac{15}{32}(64-49)$$$$=\frac{{15}^2}{32}$$$$\Rightarrow s=\frac{15}{8}$$

Commented by ajfour last updated on 28/Jan/24

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