Answer:
Let the distance of the cliff from the boat be y and height of the top of the cliff be x.
As per the statement:
From a boat on the lake, the angle of elevation to the top of a cliff is 26°1'. If the base of the cliff is 205 feet from the boat.
⇒y = 205 feet, [tex]\theta =26^{\circ} 1'[/tex]
We have to find the high is the cliff i,e x.
Using tangent ratio:
[tex]\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
See the diagram as shown below:
Opposite side = x feet
adjacent side = 205 feet
Using conversion:
[tex]1' = \frac{1}{60}^{\circ}[/tex]
then;
[tex]\theta =26^{\circ} 1' = 26\frac{1}{60} = 26.017^{\circ}[/tex]
Substitute these values we have;
[tex]\tan 26.017^{\circ} = \frac{x}{205}[/tex]
Multiply both sides by 205 we have;
[tex]x = \tan 26.017^{\circ} \cdot 205 = 0.48809992903 \cdot 205 =100.060485453[/tex] ft
Therefore, 100 ft high is the cliff(to the nearest foot)
