What is the value of \(\cos 38° \cos 52° - \sin 38° \sin 52°\)?
Answer: A
This expression is in the form of the cosine addition formula: \(\cos(A+B) = \cos A \cos B - \sin A \sin B\).
Here, A = 38° and B = 52°.
The expression is equal to \(\cos(38° + 52°) = \cos(90°)\).
The value of \(\cos 90°\) is 0.
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If \(\tan(A+B) = \sqrt{3}\) and \(\tan(A-B) = \frac{1}{\sqrt{3}}\), find A and B.
Answer: A
From the given information:
1) \(\tan(A+B) = \sqrt{3} \Rightarrow A+B = 60°\)
2) \(\tan(A-B) = \frac{1}{\sqrt{3}} \Rightarrow A-B = 30°\)
We have a system of two linear equations. Adding them gives:
2A = 90° => A = 45°.
Substituting A=45° into the first equation: 45° + B = 60° => B = 15°.
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The value of \(2\tan^2 45° + \cos^2 30° - \sin^2 60°\) is:
Answer: B
We substitute the standard trigonometric values:
\(\tan 45° = 1\)
\(\cos 30° = \frac{\sqrt{3}}{2}\)
\(\sin 60° = \frac{\sqrt{3}}{2}\)
The expression is \(2(1)^2 + (\frac{\sqrt{3}}{2})^2 - (\frac{\sqrt{3}}{2})^2 = 2(1) + \frac{3}{4} - \frac{3}{4} = 2\).
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The value of \(cosec(75°+\theta) - sec(15°-\theta) - tan(55°+\theta) + cot(35°-\theta)\) is:
Answer: B
We use the complementary angle identities: \(\sec(90°-A) = \csc A\) and \(\cot(90°-A) = \tan A\).
\(\sec(15°-\theta) = \sec(90° - (75°+\theta)) = \csc(75°+\theta)\).
\(\cot(35°-\theta) = \cot(90° - (55°+\theta)) = \tan(55°+\theta)\).
The expression becomes: \(\csc(75°+\theta) - \csc(75°+\theta) - \tan(55°+\theta) + \tan(55°+\theta) = 0\).
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An observer on top of a mountain 500 m high observes the angle of depression of two boats on the same side of the mountain to be 45° and 30°. The distance between the boats is:
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.
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The value of \(\frac{1-\tan^2 45°}{1+\tan^2 45°}\) is:
Answer: D
We know that \(\tan 45° = 1\).
Substitute this value into the expression:
\(\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\).
Alternatively, this is the formula for \(\cos(2\theta)\), so it equals \(\cos(2 \times 45°) = \cos 90° = 0\).
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The top of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with the horizontal, then the length of the wire is:
Answer: D
Imagine a right-angled triangle formed by the wire (hypotenuse), the horizontal distance, and the difference in the heights of the poles (opposite side).
The difference in height = 20 m - 14 m = 6 m.
The angle of the wire with the horizontal is 30°.
\(\sin 30° = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{Height Difference}}{\text{Wire Length}}\)
\(\frac{1}{2} = \frac{6}{\text{Wire Length}}\)
Wire Length = 6 × 2 = 12 meters.
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What is the minimum value of \(\sin^2\theta + \cos^2\theta\)?
Answer: B
This is a trick question based on the fundamental Pythagorean identity.
For any angle \(\theta\), the expression \(\sin^2\theta + \cos^2\theta\) is always equal to 1.
Therefore, its minimum value, maximum value, and average value are all 1.
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If a triangle ABC is right angled at C, then the value of \(\cos(A+B)\) is:
Answer: A
In any triangle, A + B + C = 180°.
Since the triangle is right-angled at C, we have C = 90°.
A + B + 90° = 180°
A + B = 90°.
Therefore, \(\cos(A+B) = \cos(90°) = 0\).
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The value of \(\sin(2A)\) is given by:
Answer: C
This is the standard double-angle identity for sine.
\(\sin(2A) = 2\sin A \cos A\).
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