Trigonometry Questions and Answers

Answer: A

Let the mountain be at point P, with height PQ = 500m. Let the boats be A and B.

For the farther boat (B), angle of elevation is 30°. Let its distance be d₂. \(\tan 30° = \frac{500}{d_2} \Rightarrow d_2 = 500\sqrt{3}\).

For the nearer boat (A), angle of elevation is 45°. Let its distance be d₁. \(\tan 45° = \frac{500}{d_1} \Rightarrow d_1 = 500\).

The distance between the boats is \(d_2 - d_1 = 500\sqrt{3} - 500 = 500(\sqrt{3}-1)\) m.

Answer: B

This is the cosine double angle formula, \(\cos(2A) = 1 - 2\sin^2 A\). With A = 45°, the expression is \(\cos(2 \times 45°) = \cos(90°) = 0\).

Alternatively, \(\sin 45° = \frac{1}{\sqrt{2}}\). So, \(1 - 2(\frac{1}{\sqrt{2}})^2 = 1 - 2(\frac{1}{2}) = 1 - 1 = 0\).

Answer: C

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\((\frac{a}{b})^2 + \cos^2 \theta = 1\)

\(\cos^2 \theta = 1 - \frac{a^2}{b^2} = \frac{b^2-a^2}{b^2}\)

\(\cos \theta = \sqrt{\frac{b^2-a^2}{b^2}} = \frac{\sqrt{b^2-a^2}}{b}\).

Answer: A

\[ \sin 30^{\circ} = \frac{1}{2}, \qquad \cos 60^{\circ} = \frac{1}{2} \]

\[ \left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right)^{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

The Pythagorean identity does not apply here: it needs the same angle in both terms, and 30° and 60° are different.

Answer: B

The signs of the ratios identify the quadrant.

• Quadrant I: both positive
• Quadrant II: \( \sin > 0 \), \( \cos < 0 \) ✓
• Quadrant III: both negative
• Quadrant IV: \( \sin < 0 \), \( \cos > 0 \)

So \( \theta \) lies in the second quadrant. The mnemonic "all, sine, tangent, cosine" names the ratio that is positive in each quadrant in turn.

Answer: A

We use the product-to-sum formula \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\).

Here, A = 75° and B = 15°.

The expression equals \(\cos(75°+15°) + \cos(75°-15°) = \cos 90° + \cos 60°\).

Substituting values: 0 + 1/2 = 1/2.

Answer: A

The third angle is 180° - 30° - 45° = 105°. Let the included side be 'c'. We can use the formula Area = \(\frac{c^2 \sin A \sin B}{2 \sin C}\), where C is the angle opposite side c. Here C=105°.

Area = \(\frac{(\sqrt{3}+1)^2 \sin 30° \sin 45°}{2 \sin 105°}\). This is too complex. Let's try another way. Drop a perpendicular from the vertex with angle 105 to the side c. Let it be h. Let the side c be split into x and y. Then \(\tan 30 = h/x\) and \(\tan 45 = h/y\). So \(x=h\sqrt{3}\) and \(y=h\). We have \(x+y = h(\sqrt{3}+1) = \sqrt{3}+1\). This means h=1. Area = \(1/2 \times base \times height = 1/2 \times (\sqrt{3}+1) \times 1 = \frac{\sqrt{3}+1}{2}\) cm².

Answer: C

Find a common denominator: \(\tan A (\sec A + 1)\).

Expression = \(\frac{\tan^2 A + (\sec A + 1)^2}{\tan A (\sec A + 1)}\).

Numerator = \(\tan^2 A + \sec^2 A + 2\sec A + 1\). Using \(\tan^2 A = \sec^2 A - 1\), it becomes \((\sec^2 A - 1) + \sec^2 A + 2\sec A + 1 = 2\sec^2 A + 2\sec A = 2\sec A(\sec A + 1)\).

So, Expression = \(\frac{2\sec A(\sec A + 1)}{\tan A (\sec A + 1)} = \frac{2\sec A}{\tan A} = \frac{2/\cos A}{\sin A/\cos A} = \frac{2}{\sin A} = 2\csc A\).

Answer: B

Let h be the height and x be the final distance from the lighthouse.

From the second observation: \(\tan 45° = h/x \Rightarrow 1 = h/x \Rightarrow h = x\).

From the first observation, the distance was x+100: \(\tan 30° = \frac{h}{x+100} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{h+100}\) (since x=h).

\(h+100 = h\sqrt{3} \Rightarrow 100 = h(\sqrt{3}-1) \Rightarrow h = \frac{100}{\sqrt{3}-1} = \frac{100(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{100(\sqrt{3}+1)}{3-1} = 50(\sqrt{3}+1)\) m.

Answer: B

The expression simplifies using the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\).

Numerator = \(1 - \sin^2\theta = \cos^2\theta\).

Denominator = \(1 - \cos^2\theta = \sin^2\theta\).

The expression becomes \(\frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta\).

Given \(\cot \theta = 7/8\), the value is \((\frac{7}{8})^2 = \frac{49}{64}\).