Question

The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is \(\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)\)
If the distance of the point B from the tower is 15 units, then cot α is equal to :

• A. \(\frac{6}{5}\)
• B. \(\frac{9}{5}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{7}{3}\)

Answer 1

The Correct Option is A

To solve this problem, let's assume the following: 

• The tower's height is \( h \).
• The point \( A \) is directly north of the base of the tower.
• The point \( B \) is 9 units west from \( A \).

Since the point \( B \) is 15 units away from the tower, we can apply the Pythagorean theorem to the right triangle formed by the tower, point \( A \), and point \( B \). This is because \( A \) is due north, and \( B \) is due west of \( A \).

In triangle \( ABO \):

• \( AB = 9 \) units (distance between \( A \) and \( B \))
• \( OB = 15 \) units (distance from the tower to point \( B \))
• \( OA = x \) (distance from the tower to point \( A \))

Using the Pythagorean theorem, we have:

\(x^2 + 9^2 = 15^2\)

\(x^2 + 81 = 225\)

\(x^2 = 144\)

\(x = 12\)

So, the distance \( OA \) is 12 units. Now, using the definition of cotangent, which is the adjacent over the opposite side in the right triangle:

The angle of elevation from \( A \) can be written in terms of its tangent:

\(\tan \alpha = \frac{h}{12}\)

The given angle at \( B \) is:

\(\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)\)

Since the secant of this angle gives the distance from \( O \) to \( B \):

\(\cos(\theta) = \frac{3}{\sqrt{13}}\)

Using trigonometric identities, the sine of this angle gives the height \( h \):

Calculate \( \sin \theta \):

\(\sin \theta = \sqrt{1 - \left(\frac{3}{\sqrt{13}}\right)^2} = \frac{2}{\sqrt{13}}\)

The cotangent of angle \( \alpha \) is:

\(\cot \alpha = \frac{12}{h}\)

Since \( \sin \theta = \frac{h}{15} \), we have:

\(h = 15 \sin \theta = 15 \times \frac{2}{\sqrt{13}}\)\)

\(h = \frac{30}{\sqrt{13}}\)

Finally, plug \( h \) back in to find \( \cot \alpha \):

\(\cot \alpha = \frac{12}{\left(\frac{30}{\sqrt{13}}\right)} = \frac{12 \sqrt{13}}{30} = \frac{6}{5}\)

Therefore, the cotangent of angle \( \alpha \) is \(\frac{6}{5}\), making the correct answer:

\(\frac{6}{5}\)

Answer 2

Apply Pythagoras Theorem in NAB,
\(NA = \sqrt{15^2 - 9^2}\)
\(NA=12\)
\(\frac{h}{15} = \tan \theta = \frac{2}{3}\)
\(h = 10\ units\)
\(\cot \alpha = \frac{12}{10}\)
\(\cot \alpha = \frac{6}{5}\)
So, the correct option is (A): \(\frac{6}{5}\)