In a right-angled triangle ABC, right-angled at B, if \(\tan A = 1\), then the value of \(2 \sin A \cos A\) is:
Answer: B
If \(\tan A = 1\), then A = 45°.
The expression \(2 \sin A \cos A\) is the double angle identity for sine, \(\sin(2A)\).
So, we need to find \(\sin(2 \times 45°) = \sin(90°)\).
The value of \(\sin 90°\) is 1.
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The value of \(9 \sec^2 A - 9 \tan^2 A\) is:
Answer: B
Take out the common factor and use the identity.
\[ 9(\sec^{2}A - \tan^{2}A) = 9 \times 1 = 9 \]
The identity \( \sec^{2}A - \tan^{2}A = 1 \) comes from dividing \( \sin^{2}A + \cos^{2}A = 1 \) through by \( \cos^{2}A \).
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From a point on the ground, 20 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?
Answer: B
\[ \tan 60^{\circ} = \frac{h}{20} \]
\[ h = 20\sqrt{3} \approx 34.6 \text{ m} \]
At 60° the tower is \( \sqrt{3} \) times the ground distance - taller than 20 m, which rules out two of the options at once.
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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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A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is:
Answer: C
Be careful with the angle. The angle is with the *wall*, not the ground.
Let h be the height of the wall (adjacent side to the 60° angle) and L be the length of the ladder (hypotenuse).
\(\cos(\text{angle}) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
\(\cos 60° = \frac{h}{15}\)
\(\frac{1}{2} = \frac{h}{15}\)
h = 15 / 2 = 7.5 meters.
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If \(\sin(A - B) = 1/2\) and \(\cos(A + B) = 1/2\), the value of B is:
Answer: A
From the given information:
1) \(\sin(A - B) = 1/2 \Rightarrow A - B = 30°\)
2) \(\cos(A + B) = 1/2 \Rightarrow A + B = 60°\)
We have two equations. To find B, subtract the first equation from the second:
(A + B) - (A - B) = 60° - 30°
2B = 30°
B = 15°.
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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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