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Question Number 152949 by Riyoziyot last updated on 03/Sep/21

Answered by mr W last updated on 03/Sep/21

say AB=AC=1 AD=CD=(1/(2 cos 42)) BC=2 cos 66 ∠BCD=66−42=24° ((sin (x+24))/(sin x))=((BC)/(CD))=4 cos 66 cos 42 cos 24+((sin 24)/(tan x))=4 cos 66 cos 42 tan x=((sin 24)/(4 cos 66 cos 42−cos 24)) x=tan^(−1) ((sin 24)/(4 cos 66 cos 42−cos 24))=54°

$${say}\:{AB}={AC}=\mathrm{1} \\ $$$${AD}={CD}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{42}} \\ $$$${BC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{66} \\ $$$$\angle{BCD}=\mathrm{66}−\mathrm{42}=\mathrm{24}° \\ $$$$\frac{\mathrm{sin}\:\left({x}+\mathrm{24}\right)}{\mathrm{sin}\:{x}}=\frac{{BC}}{{CD}}=\mathrm{4}\:\mathrm{cos}\:\mathrm{66}\:\mathrm{cos}\:\mathrm{42} \\ $$$$\mathrm{cos}\:\mathrm{24}+\frac{\mathrm{sin}\:\mathrm{24}}{\mathrm{tan}\:{x}}=\mathrm{4}\:\mathrm{cos}\:\mathrm{66}\:\mathrm{cos}\:\mathrm{42} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:\mathrm{24}}{\mathrm{4}\:\mathrm{cos}\:\mathrm{66}\:\mathrm{cos}\:\mathrm{42}−\mathrm{cos}\:\mathrm{24}} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{sin}\:\mathrm{24}}{\mathrm{4}\:\mathrm{cos}\:\mathrm{66}\:\mathrm{cos}\:\mathrm{42}−\mathrm{cos}\:\mathrm{24}}=\mathrm{54}° \\ $$

Commented by Tawa11 last updated on 03/Sep/21

Weldone sir.

$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$

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