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Question Number 60283 by mr W last updated on 19/May/19

Commented by mr W last updated on 19/May/19

∠B=15°, ∠C=30°, BD=DC  find x=?

$$\angle{B}=\mathrm{15}°,\:\angle{C}=\mathrm{30}°,\:{BD}={DC} \\ $$$${find}\:{x}=? \\ $$

Answered by ajfour last updated on 19/May/19

Let BD=CD=1  ((sin x)/1)=((sin 15°)/(AD))    &  ((sin (135°−x))/1)=((sin 30°)/(AD))  ⇒ sin (135°−x)=((sin x)/(2sin 15°))  ⇒  2sin 15°(((cos x+sin x)/(√2)))=sin x       2((((√3)−1)/(2(√2))))(((cos x+sin x)/(√2)))=sin x  ⇒   cot x+1 = (2/((√3)−1))  ⇒   cot x= (√3)    ⇒ x=30° .

$$\mathrm{Let}\:\mathrm{BD}=\mathrm{CD}=\mathrm{1} \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}}=\frac{\mathrm{sin}\:\mathrm{15}°}{\mathrm{AD}}\:\:\:\:\&\:\:\frac{\mathrm{sin}\:\left(\mathrm{135}°−\mathrm{x}\right)}{\mathrm{1}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{AD}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\mathrm{135}°−\mathrm{x}\right)=\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2sin}\:\mathrm{15}°} \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:\mathrm{15}°\left(\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\sqrt{\mathrm{2}}}\right)=\mathrm{sin}\:\mathrm{x} \\ $$$$\:\:\:\:\:\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\left(\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\sqrt{\mathrm{2}}}\right)=\mathrm{sin}\:\mathrm{x} \\ $$$$\Rightarrow\:\:\:\mathrm{cot}\:\mathrm{x}+\mathrm{1}\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}−\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\mathrm{cot}\:\mathrm{x}=\:\sqrt{\mathrm{3}}\:\:\:\:\Rightarrow\:\boldsymbol{\mathrm{x}}=\mathrm{30}°\:. \\ $$

Commented by mr W last updated on 19/May/19

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

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