Question

question 5 canada’s highest waterfall is della falls on vancouver island. an observer standing at the same level as the base of the falls views the top of the falls at an angle of elevation of 58°. when the observer moves 31 m closer to the base of the falls the angle of elevation increases to 61°. find the height of della falls. answer to the nearest metre. > show answer the answer is: 440 m is your answer correct? yes no

Explanation:

Step1: Define variables

Let \( h \) be the height of the falls, \( x \) be the distance from the observer's new position (after moving 31 m closer) to the base of the falls. Then the original distance from the observer to the base is \( x + 31 \).

Step2: Set up trigonometric equations

Using the tangent function for the two angles of elevation:
For the angle \( 58^\circ \): \( \tan(58^\circ)=\frac{h}{x + 31} \), so \( h=(x + 31)\tan(58^\circ) \)
For the angle \( 61^\circ \): \( \tan(61^\circ)=\frac{h}{x} \), so \( h = x\tan(61^\circ) \)

Step3: Equate the two expressions for \( h \)

\( x\tan(61^\circ)=(x + 31)\tan(58^\circ) \)
Expand the right - hand side: \( x\tan(61^\circ)=x\tan(58^\circ)+31\tan(58^\circ) \)
Subtract \( x\tan(58^\circ) \) from both sides: \( x\tan(61^\circ)-x\tan(58^\circ)=31\tan(58^\circ) \)
Factor out \( x \): \( x(\tan(61^\circ)-\tan(58^\circ)) = 31\tan(58^\circ) \)

Step4: Solve for \( x \)

We know that \( \tan(61^\circ)\approx1.8040 \), \( \tan(58^\circ)\approx1.6003 \)
\( x=\frac{31\tan(58^\circ)}{\tan(61^\circ)-\tan(58^\circ)}=\frac{31\times1.6003}{1.8040 - 1.6003}=\frac{49.6093}{0.2037}\approx243.54 \)

Step5: Solve for \( h \)

Using \( h = x\tan(61^\circ) \), substitute \( x\approx243.54 \) and \( \tan(61^\circ)\approx1.8040 \)
\( h=243.54\times1.8040\approx440 \)

Answer:

\( 440 \) metres