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Question Number 206804 by mr W last updated on 25/Apr/24

Answered by mr W last updated on 26/Apr/24

Commented by mr W last updated on 26/Apr/24

$$\frac{3y}{5y}\times \frac{a}{b}\times \frac{c+d}{d}=1$$$$\Rightarrow \frac{c}{d}=\frac{5b-3a}{3a}...\left(i\right)$$$$\frac{x}{3x}\times \frac{d}{c}\times \frac{b+a}{a}=1$$$$\Rightarrow \frac{c}{d}=\frac{a+b}{3a}...\left(ii\right)$$$$\Rightarrow a+b=5b-3a$$$$\Rightarrow a=b$$$$\frac{BA}{BC}=\frac{b}{a}=1\Rightarrow BA=BC$$$$\Rightarrow \mathrm{\Delta }ABD\equiv \mathrm{\Delta }CBD$$$$\Rightarrow \mathrm{\angle }A=\mathrm{\angle }C=\frac{180-2\times 35}{2}=55°✓$$

Answered by A5T last updated on 26/Apr/24

Commented by mr W last updated on 26/Apr/24

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Commented by A5T last updated on 26/Apr/24

$$\frac{sin\theta }{3x}=\frac{sin35}{3y}\Rightarrow sin\theta =\frac{xsin35}{y}$$$$\frac{sin\theta }{AB}=\frac{sin70}{8y}\Rightarrow AB=\frac{4x}{cos35}$$$${BC}^2=16x^2+\frac{16x^2}{{cos}^{2}35}-32x^2=\frac{16x^2}{{cos}^{2}35}-16x^2$$$$\Rightarrow BC=4x\sqrt{\frac{1}{{cos}^{2}35}-1}=\frac{4xsin35}{cos35}$$$$\frac{sin(35+?)}{AB}=\frac{sin35}{BC}\Rightarrow \frac{sin(35+?)cos35}{4x}=\frac{cos35sin35}{4xsin35}$$$$sin(35+?)=1\Rightarrow ?=55°$$

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