Question Number 72414 by aliesam last updated on 28/Oct/19 | ||
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Commented by aliesam last updated on 28/Oct/19 | ||
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$${a},{b}\:{and}\:{d}\:{are}\:{centres}\:{of}\:{three}\:{circles}\: \\ $$$${find}\: \\ $$$$\frac{{AC}}{{AB}} \\ $$ | ||
Answered by mr W last updated on 28/Oct/19 | ||
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$${let}\:{AC}={a},\:{AB}={BD}={b} \\ $$$${a}\:\mathrm{sin}\:\mathrm{30}°={b}\:\mathrm{sin}\:\mathrm{60}° \\ $$$$\frac{{AC}}{{AB}}=\frac{{a}}{{b}}=\frac{\mathrm{sin}\:\mathrm{60}°}{\mathrm{sin}\:\mathrm{30}°}=\sqrt{\mathrm{3}} \\ $$ | ||
Commented by aliesam last updated on 28/Oct/19 | ||
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$${god}\:{bless}\:{you}\:{sir} \\ $$ | ||
Commented by mr W last updated on 28/Oct/19 | ||
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