Trigonometry Questions and Answers

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.

Answer: B

Substitute the standard values:

\(2(\frac{1}{2})^2 + 3(\frac{1}{\sqrt{2}})^2 - 3(\frac{1}{2})^2 + 2(0)^2\)

= \(2(\frac{1}{4}) + 3(\frac{1}{2}) - 3(\frac{1}{4}) + 0\)

= \(\frac{1}{2} + \frac{3}{2} - \frac{3}{4}\)

= \(\frac{4}{2} - \frac{3}{4} = 2 - \frac{3}{4} = \frac{8-3}{4} = \frac{5}{4}\).

Answer: B

Let \(x = \sin A\). Then \(\csc A = 1/x\). We are given \(x + 1/x = 2\). This is only possible if x=1. So, \(\sin A = 1\).

Then \(\csc A = 1/\sin A = 1\).

The expression becomes \(\sin^2 A + \csc^2 A = 1^2 + 1^2 = 1 + 1 = 2\).

Alternatively, square the given equation: \((\sin A + \csc A)^2 = 2^2\).

\(\sin^2 A + \csc^2 A + 2\sin A \csc A = 4\).

Since \(\sin A \csc A = 1\), we have \(\sin^2 A + \csc^2 A + 2 = 4\).

\(\sin^2 A + \csc^2 A = 2\).

Answer: A

\[ \tan\theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{\sqrt{3}\,h} = \frac{1}{\sqrt{3}} \]

\[ \theta = 30^{\circ} \]

A shadow longer than the object always means the sun is below 45°.

Answer: C

We find a common denominator, which is \(\sin\theta(1+\cos\theta)\).

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

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

Using \(\sin^2\theta + \cos^2\theta = 1\), the numerator becomes \(1 + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1+\cos\theta)\).

= \(\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} = 2\csc\theta\).

Answer: A

Let H be the total height of the tower. The height of the tower above the observer's eye level is h'.

The distance is d = 20√3 m.

\(\tan 30° = \frac{h'}{d} = \frac{h'}{20\sqrt{3}}\)

\(\frac{1}{\sqrt{3}} = \frac{h'}{20\sqrt{3}} \Rightarrow h' = 20\) m.

The total height of the tower H = h' + height of observer = 20 + 1.6 = 21.6 m.

Answer: C

From the given equation, \(\tan A = 4/5\).

Divide the numerator and denominator of the expression by \(\cos A\):

\(\frac{5\tan A - 3}{5\tan A + 2}\)

Substitute the value of \(\tan A\):

\(\frac{5(4/5) - 3}{5(4/5) + 2} = \frac{4-3}{4+2} = \frac{1}{6}\).