Question

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, how high is the cliff (to the nearest foot)?

Answer 1

We use the tangent for this

tan 26 1   = opposite / adjacent =  h / 205   where h = height of cliff

h = 205 tan 26 1   =    100 feet

Answer 2

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, $\theta =26^{\circ} 1'$

We have to find the high is the cliff i,e x.

Using tangent ratio:

$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$

See the diagram as shown below:

Opposite side = x feet

adjacent side = 205 feet

Using conversion:

$1' = \frac{1}{60}^{\circ}$

then;

$\theta =26^{\circ} 1' = 26\frac{1}{60} = 26.017^{\circ}$

Substitute these values we have;

$\tan 26.017^{\circ} = \frac{x}{205}$

Multiply both sides by 205 we have;

$x = \tan 26.017^{\circ} \cdot 205 = 0.48809992903 \cdot 205 =100.060485453$ ft

Therefore, 100 ft high is the cliff(to the nearest foot)