Trigonometry Questions and Answers

Answer: B

Use the complementary-angle identity \( \cos(90^{\circ} - \theta) = \sin\theta \).

\[ \cos 72^{\circ} = \cos(90^{\circ} - 18^{\circ}) = \sin 18^{\circ} \]

\[ \frac{\sin 18^{\circ}}{\sin 18^{\circ}} = 1 \]

Neither value has to be computed - only recognised as equal.

Answer: B

Let the height be h and the shadow length be s. We are given h = s.

The angle of elevation θ is given by \(\tan \theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{s}\).

Since h = s, \(\tan \theta = 1\).

The angle whose tangent is 1 is 45°.

Answer: C

We use the identity \(\cot \theta = \tan(90° - \theta)\).

So, \(\cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A)\).

The equation becomes \(\tan 2A = \tan(108° - A)\).

This implies \(2A = 108° - A\).

3A = 108°

A = 36°.

Answer: B

Let H be the height of the tower and h=25m be the height of the cliff. Let d be the horizontal distance between them. Let θ be the angle.

Angle of depression to the foot: \(\tan \theta = \frac{h}{d} = \frac{25}{d}\).

Angle of elevation to the top: \(\tan \theta = \frac{H-h}{d} = \frac{H-25}{d}\).

Since the angles are equal, their tangents are equal.

\(\frac{25}{d} = \frac{H-25}{d} \Rightarrow 25 = H - 25 \Rightarrow H = 50\) m.

Answer: B

This is a fundamental Pythagorean identity, derived by dividing \(\sin^2 A + \cos^2 A = 1\) by \(\sin^2 A\).

\(\frac{\sin^2 A}{\sin^2 A} + \frac{\cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A}\)

\(1 + \cot^2 A = \csc^2 A\).

Answer: B

Correct answer: 25/3

Factor the 3 out and use a Pythagorean identity rather than finding \(\tan X\) first.

\[3 + 3\tan^2 X = 3\left(1 + \tan^2 X\right) = 3\sec^2 X\]

Since \(\cos X = \tfrac{3}{5}\), we have \(\sec X = \tfrac{5}{3}\), so

\[3\sec^2 X = 3 \times \frac{25}{9} = \frac{25}{3}\]

The same result follows the long way: \(\cos X = \tfrac{3}{5}\) gives a 3–4–5 triangle with \(\tan X = \tfrac{4}{3}\), and \(3 + 3\left(\tfrac{16}{9}\right) = 3 + \tfrac{16}{3} = \tfrac{25}{3}\). Stopping at \(\tfrac{16}{3}\) omits the leading 3.

Answer: C

First, we can factor out the 3 from the expression: \(3(1 + \tan^2 X)\).

Using the Pythagorean identity \(1 + \tan^2 X = \sec^2 X\), the expression simplifies to \(3 \sec^2 X\).

We are given \(\cos X = 3/5\). Since \(\sec X = 1/\cos X\), we have \(\sec X = 5/3\).

Now substitute this value into the expression:

Value = \(3 \times (\frac{5}{3})^2 = 3 \times \frac{25}{9} = \frac{25}{3}\).

Answer: C

This is the sine addition formula, \(\sin(A+B) = \sin A \cos B + \cos A \sin B\).

The expression is equal to \(\sin(30°+60°) = \sin(90°) = 1\).

Alternatively, substituting values: \((1/2)(1/2) + (\sqrt{3}/2)(\sqrt{3}/2) = 1/4 + 3/4 = 4/4 = 1\).

Answer: A

The ladder is the hypotenuse (20 m) and the height on the wall is the opposite side (10 m).

Let θ be the angle with the ground.

\(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{10}{20} = \frac{1}{2}\).

The angle whose sine is 1/2 is 30°.

Answer: B

Sine and cosine are cofunctions:

\[ \sin(90^{\circ} - A) = \cos A \]

The two angles are complementary, and in a right triangle the side opposite one is adjacent to the other - which is where the identity comes from.