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Question Number 125195 by Eric002 last updated on 09/Dec/20

Answered by mr W last updated on 09/Dec/20

$$\frac{AD}{\sin 58}=\frac{DC}{\sin x}...\left(i\right)$$$$\frac{AD}{\sin \mathrm{\angle }B}=\frac{BD}{\sin 32}$$$$\mathrm{\angle }B=180-58-x+32=90-x$$$$\frac{AD}{\cos x}=\frac{DC}{\sin 32}...\left(ii\right)$$$$\left(ii\right)\mathbf{÷}\left(i\right):$$$$\frac{\cos x}{\sin 58}=\frac{\sin 32}{\sin x}$$$$\Rightarrow \sin x\cos x=\sin 32\cos 32=\sin 58\cos 58$$$$\Rightarrow x=32°or58°$$

Commented by Eric002 last updated on 09/Dec/20

$$welldone$$

Commented by mr W last updated on 09/Dec/20

Commented by mr W last updated on 09/Dec/20

$$thisshowswhytwosolutionsexist.$$

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