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Question-58313




Question Number 58313 by mr W last updated on 21/Apr/19
Commented by tanmay last updated on 21/Apr/19
Commented by tanmay last updated on 21/Apr/19
wait sir trying to find x...
$${wait}\:{sir}\:{trying}\:{to}\:{find}\:{x}… \\ $$
Answered by mr W last updated on 21/Apr/19
Commented by mr W last updated on 21/Apr/19
Method 1 using trigonometry:  ((AC)/(AD))=((sin (x+20))/(sin x))  ((AC)/(AB))=((sin 80)/(sin 40))  since AB=AD,  ⇒((sin (x+20))/(sin x))=((sin 80)/(sin 40))=2 cos 40=2 sin 50=2 sin (30+20)=cos 20+(√3) sin 20  ⇒cos 20+((sin 20)/(tan x))=cos 20+(√3) sin 20  ⇒(1/(tan x))=(√3)  ⇒tan x=(1/( (√3)))  ⇒x=30°
$${Method}\:\mathrm{1}\:{using}\:{trigonometry}: \\ $$$$\frac{{AC}}{{AD}}=\frac{\mathrm{sin}\:\left({x}+\mathrm{20}\right)}{\mathrm{sin}\:{x}} \\ $$$$\frac{{AC}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{40}} \\ $$$${since}\:{AB}={AD}, \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\left({x}+\mathrm{20}\right)}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{40}}=\mathrm{2}\:\mathrm{cos}\:\mathrm{40}=\mathrm{2}\:\mathrm{sin}\:\mathrm{50}=\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{30}+\mathrm{20}\right)=\mathrm{cos}\:\mathrm{20}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{20} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{20}+\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{tan}\:{x}}=\mathrm{cos}\:\mathrm{20}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{20} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:{x}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\mathrm{30}° \\ $$
Answered by mr W last updated on 21/Apr/19
Commented by mr W last updated on 21/Apr/19
Method 2 using only basic geometry:  construct AE with ∠BAE=20°  AB=AE=AD=ED=EC  ∠ECD=∠EDC=70°  ⇒x=70°−40°=30°
$${Method}\:\mathrm{2}\:{using}\:{only}\:{basic}\:{geometry}: \\ $$$${construct}\:{AE}\:{with}\:\angle{BAE}=\mathrm{20}° \\ $$$${AB}={AE}={AD}={ED}={EC} \\ $$$$\angle{ECD}=\angle{EDC}=\mathrm{70}° \\ $$$$\Rightarrow{x}=\mathrm{70}°−\mathrm{40}°=\mathrm{30}° \\ $$
Commented by tanmay last updated on 21/Apr/19
thsnk yousir...
$${thsnk}\:{yousir}… \\ $$
Commented by salahahmed last updated on 22/Apr/19
Why is ED=EC ?
$$\mathrm{Why}\:\mathrm{is}\:{ED}={EC}\:? \\ $$
Commented by mr W last updated on 22/Apr/19
∠CAE=∠ACE=40°  ⇒EC=EA=ED
$$\angle{CAE}=\angle{ACE}=\mathrm{40}° \\ $$$$\Rightarrow{EC}={EA}={ED} \\ $$
Commented by salahahmed last updated on 23/Apr/19
Thank you sir

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