$Trigonometry Question on 3D shape (10th grade)$

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$Trigonometry Question on 3D shape (10th grade)$

Trigonometry

Thomas Marteau

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Savionf

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Low bounty.
- Thomas Marteau
  
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  Sorry brother
Blue

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Is there any additional information in this diagram?
- Thomas Marteau
  
  0
  
  Yes, angles DBC and DBA are right angle, 90 degrees.

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Blue

+1

Is there any additional information in this diagram?

Thomas Marteau

0

Yes, angles DBC and DBA are right angle, 90 degrees.
Thomas Marteau

0

Yes, angles DBC and DBA are right angle, 90 degrees.

Answer 1

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We have

\[\cot 35=\frac{AB}{x} \Rightarrow AB=x\cot 35,\]
and

\[\cot 40=\frac{BC}{x} \Rightarrow BC=x\cot 40.\]
By law of cosines

\[AC^2=AB^2+BC^2-2AB\cdot BC \cos 45\]
\[\Rightarrow 20^2=x^2\cot^2 35 +x^2\cot^2 40 -2 x^2 \cot 34 \cot 45=x^2(\cot^2 35+\cot^2 40-2\cos 45)= 3.841 x^2.\]
Hence
\[x^2=\frac{20^2}{3.841}=104.1395\]
\[\Rightarrow x=10.20 cm.\]

Erdos

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Erdos

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Please consider offering slightly higher bounties for your future questions.
- Thomas Marteau
  
  +1
  
  thanks so much man it’s my first time i’ll do so next time!

$Trigonometry Question on 3D shape (10th grade)$

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