Answer:
307 ft
Step-by-step explanation:
We can model this problem as two right triangles formed by the observer, the base of the tower, and the top of the tower.
As we have been given the angles of elevation and the length of the sides adjacent the angles, and we wish to find the length of the sides opposite the angles (height of the tower), we can use the tangent trigonometric ratio.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Tangent trigonometric ratio}}\\\\\sf \tan(\theta)=\dfrac{O}{A}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$O$ is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\end{array}}[/tex]
Let x be the horizontal distance between the visitor and the bridge after sailing 400 ft closer.
For the first right triangle, where the visitor estimates that the angle of elevation is 28°:
Substitute these values into the tangent ratio and solve for x:
[tex]\tan 28^{\circ}=\dfrac{h}{400+x}\\\\\\\tan 28^{\circ}(400+x)=h\\\\\\400\tan 28^{\circ}+x\tan 28^{\circ}=h\\\\\\x\tan 28^{\circ}=h-400\tan 28^{\circ}\\\\\\x=\dfrac{h-400\tan 28^{\circ}}{\tan 28^{\circ}}[/tex]
For the second right triangle, where the visitor estimates that the angle of elevation is 60° after sailing 400 ft closer to the bridge:
Substitute these values into the tangent ratio and solve for x:
[tex]\tan 60^{\circ}=\dfrac{h}{x}\\\\\\x=\dfrac{h}{\tan 60^{\circ}}\\\\\\x=\dfrac{h}{\sqrt{3}}[/tex]
To find the height (h), set the two equations equal to each other and solve for h:
[tex]\dfrac{h-400\tan 28^{\circ}}{\tan 28^{\circ}}=\dfrac{h}{\sqrt{3}} \\\\\\\\ \dfrac{h}{\tan 28^{\circ}}-\dfrac{400\tan 28^{\circ}}{\tan 28^{\circ}}=\dfrac{h}{\sqrt{3}} \\\\\\\\ \dfrac{h}{\tan 28^{\circ}}-400=\dfrac{h}{\sqrt{3}} \\\\\\\\ \dfrac{h}{\tan 28^{\circ}}-\dfrac{h}{\sqrt{3}}=400 \\\\\\\\ \dfrac{h\sqrt{3}}{\sqrt{3}\tan 28^{\circ}}-\dfrac{h\tan 28^{\circ}}{\sqrt{3}\tan 28^{\circ}}=400 \\\\\\\\ \dfrac{h\sqrt{3}-h\tan 28^{\circ}}{\sqrt{3}\tan 28^{\circ}}=400[/tex]
[tex]h\sqrt{3}-h\tan 28^{\circ}=400\sqrt{3}\tan 28^{\circ} \\\\\\\\ h(\sqrt{3}-\tan 28^{\circ})=400\sqrt{3}\tan 28^{\circ} \\\\\\\\ h=\dfrac{400\sqrt{3}\tan 28^{\circ}}{\sqrt{3}-\tan 28^{\circ}} \\\\\\\\ h=306.89527795...\\\\\\h=307\; \sf ft\;(nearest\;integer)[/tex]
Therefore, the height of the tower is approximately:
[tex]\Large\boxed{\boxed{307\; \sf ft}}[/tex]
