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In triangle ABC (see Figure 1.4),  BCA = 90◦, and D is the foot of
the perpendicular line segment from C to segment AB. Given that |AB| = x and
 A = θ, express all the lengths of the segments in Figure 1.4 in terms of x and θ.

 Aug 16, 2019
 #1
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In triangle ABC (see Figure 1.4), \(\angle BCA = 90^\circ\) , and D is the foot of
the perpendicular line segment from C to segment AB. Given that |AB| = x and

\(\angle A = \theta\), express all the lengths of the segments in Figure 1.4 in terms of x and \(\theta\).

 

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{|CB|} &=& \mathbf{x\cdot \sin(\theta)} \\ (2) & \mathbf{|CA|} &=& \mathbf{x\cdot \cos(\theta)} \\\\ & \sin(\theta) &=& \dfrac{|CD|}{|CA|} \\ & |CD| &=& |CA|\cdot \sin(\theta) \\ (3)& \mathbf{|CD|} &=& \mathbf{x\cdot \cos(\theta)\cdot \sin(\theta) } \\\\ & \cos(\theta) &=& \dfrac{|AD|}{|CA|} \\ & |AD| &=& |CA|\cdot \cos(\theta) \\ & |AD| &=& x\cdot \cos(\theta)\cdot \cos(\theta) \\ (4)& \mathbf{|AD|} &=& \mathbf{x\cdot \cos^2(\theta)} \\\\ & |DB| &=& x-|AD| \\ & |DB| &=& x-x\cdot \cos^2(\theta) \\ & |DB| &=& x\cdot\left(1-\cos^2(\theta)\right) \quad | \quad \sin^2(\theta)+cos^2(\theta)=1 \\ (5)& \mathbf{|DB|} &=& \mathbf{x\cdot\sin^2(\theta)} \\ \hline \end{array}\)

 

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 Aug 16, 2019

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