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4. The angle of elevation of the Top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y 40m vertically above X the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX(use root 3= 1.73)

SARAHann  Mar 10, 2017
 #1
avatar+19653 
0

4. The angle of elevation of the Top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y 40m vertically above X the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX(use root 3= 1.73)

 

Let x = PQ (height )
Let y = PX (distance)

 

\(\begin{array}{|rcll|} \hline \tan(60^{\circ}) = \frac{x}{y} &=& \sqrt{3} \qquad & | \qquad \tan(60^{\circ}) = \sqrt{3} \\ y &=& \frac{x}{\sqrt{3}} \\\\ \tan(45^{\circ}) = \frac{x-40}{y} &=& 1 \qquad & | \qquad \tan(45^{\circ}) = 1 \\ x-40 &=& y \\\\ x-40 &=& \frac{x}{\sqrt{3}} \\ \dots \\ x &=& \frac{ 40\cdot \sqrt{3} } {\sqrt{3}-1} \qquad & | \qquad \sqrt{3} = 1.73 \\ &=& \frac{ 40\cdot 1.73 } {1.73-1} \\ &=& \frac{ 69.2 } {0.73} \\ \mathbf{x} & \mathbf{=} & \mathbf{94.79\ m} \\\\ y &=& \frac{x}{\sqrt{3}} \\ &=& \frac{94.79}{\sqrt{3}} \qquad & | \qquad \sqrt{3} = 1.73 \\ &=& \frac{94.79}{1.73} \\ \mathbf{y} & \mathbf{=} & \mathbf{54.79\ m} \\\\ \hline \end{array} \)

 

laugh

heureka  Mar 10, 2017
 #2
avatar+302 
0

Tysm fr the answer!!😊😊

SARAHann  Mar 10, 2017

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