Trigonometry Questions and Answers

Answer: B

The angle of depression from the top of the lighthouse to the boat is equal to the angle of elevation from the boat to the top of the lighthouse.

Let the distance of the boat from the foot of the lighthouse be 'd'.

\(\tan 30° = \frac{\text{height}}{\text{distance}} = \frac{100}{d}\)

\(\frac{1}{\sqrt{3}} = \frac{100}{d}\)

d = 100√3 meters.

Answer: C

\[ \cos 0^{\circ} = 1, \qquad \sin 90^{\circ} = 1, \qquad \tan 0^{\circ} = 0 \]

\[ 1 + 1 + 0 = 2 \]

Answer: A

We know that \(\sec^2 \theta - \tan^2 \theta = 1\), which can be factored as \((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\).

Given \(\sec \theta + \tan \theta = x\), we have \(\sec \theta - \tan \theta = \frac{1}{x}\).

Now we have a system of two linear equations:

1) \(\sec \theta + \tan \theta = x\)

2) \(\sec \theta - \tan \theta = 1/x\)

Adding the two equations gives: \(2\sec \theta = x + \frac{1}{x} = \frac{x^2+1}{x}\).

Therefore, \(\sec \theta = \frac{x^2+1}{2x}\).

Answer: C

For an expression of the form \(a\sin \theta + b\cos \theta\), the maximum value is \(\sqrt{a^2+b^2}\) and the minimum value is \(-\sqrt{a^2+b^2}\).

Here, a = 3 and b = 4.

Maximum value = \(\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\).

Answer: B

We use the complementary angle identity \(\tan(90° - \theta) = \cot \theta\), and the fact that \(\tan \theta \cdot \cot \theta = 1\).

The expression can be paired up: \((\tan 1° \tan 89°) (\tan 2° \tan 88°) ...\)

Since \(\tan 89° = \tan(90°-1°) = \cot 1°\), the first pair is \(\tan 1° \cot 1° = 1\).

All pairs will multiply to 1. The middle term is \(\tan 45°\), which is also 1.

The final product is 1.

Answer: B

From the given equation, we can write:

\(\sin \theta = 1 - \sin^2 \theta\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we know that \(1 - \sin^2 \theta = \cos^2 \theta\).

Therefore, we have \(\sin \theta = \cos^2 \theta\).

Now we evaluate the expression \(\cos^2 \theta + \cos^4 \theta\). We can write \(\cos^4 \theta\) as \((\cos^2 \theta)^2\).

Substituting \(\cos^2 \theta = \sin \theta\), the expression becomes:

\(\sin \theta + (\sin \theta)^2 = \sin \theta + \sin^2 \theta\)

From the original given equation, we know that \(\sin \theta + \sin^2 \theta = 1\).

So, the value of the expression is 1.

Answer: A

This expression is in the form of the tangent subtraction formula: \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\).

Here, A = 60° and B = 30°.

So, the expression is equal to \(\tan(60° - 30°) = \tan 30°\).

Answer: D

Let the length of the ladder be L (the hypotenuse) and the distance from the wall be D (the adjacent side).

We use the cosine function: \(\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}\).

\(\cos 60° = \frac{4.6}{L}\)

We know \(\cos 60° = 1/2\).

\(\frac{1}{2} = \frac{4.6}{L}\)

L = 4.6 × 2 = 9.2 meters.

Answer: A

We can write 75° as the sum of two standard angles, 45° + 30°.

Using the sine addition formula: \(\sin(A+B) = \sin A \cos B + \cos A \sin B\).

\(\sin(45°+30°) = \sin 45° \cos 30° + \cos 45° \sin 30°\)

= \((\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}})(\frac{1}{2})\)

= \(\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}\).

Answer: D

If \(\sin \theta = \cos \theta\), dividing both sides by \(\cos \theta\) gives \(\tan \theta = 1\). This means \(\theta = 45°\).

Now we evaluate the expression: \(2\tan 45° + \cos^2 45°\).

= \(2(1) + (\frac{1}{\sqrt{2}})^2\)

= \(2 + \frac{1}{2} = \frac{5}{2}\).