Find the value of \(2\cos^2 60° + 3\sin^2 45° - 3\sin^2 30° + 2\cos^2 90°\).
Answer: B
Substitute the standard values:
\(2(\frac{1}{2})^2 + 3(\frac{1}{\sqrt{2}})^2 - 3(\frac{1}{2})^2 + 2(0)^2\)
= \(2(\frac{1}{4}) + 3(\frac{1}{2}) - 3(\frac{1}{4}) + 0\)
= \(\frac{1}{2} + \frac{3}{2} - \frac{3}{4}\)
= \(\frac{4}{2} - \frac{3}{4} = 2 - \frac{3}{4} = \frac{8-3}{4} = \frac{5}{4}\).
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If \(\sin A + \csc A = 2\), then the value of \(\sin^2 A + \csc^2 A\) is:
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 \(\sin^2 A + \csc^2 A + 2 = 4\).
\(\sin^2 A + \csc^2 A = 2\).
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The shadow of a tower is √3 times its height. The angle of elevation of the sun is:
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°.
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The value of \(\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta}\) is:
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 \(1 + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1+\cos\theta)\).
= \(\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} = 2\csc\theta\).
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An observer 1.6 m tall is 20√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:
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.
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If \(5\tan A = 4\), the value of \(\frac{5\sin A - 3\cos A}{5\sin A + 2\cos A}\) is:
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 \(\tan A\):
\(\frac{5(4/5) - 3}{5(4/5) + 2} = \frac{4-3}{4+2} = \frac{1}{6}\).
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What is the value of \(\tan 75° + \cot 75°\)?
Answer: C
First, find the values of \(\tan 75°\) and \(\cot 75°\).
\(\tan 75° = \tan(45°+30°) = \frac{\tan 45 + \tan 30}{1 - \tan 45 \tan 30} = \frac{1+1/\sqrt{3}}{1-1/\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}\).
\(\cot 75° = 1/\tan 75° = \frac{1}{2+\sqrt{3}} = 2-\sqrt{3}\).
The sum is \((2+\sqrt{3}) + (2-\sqrt{3}) = 4\).
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The value of \(2\sin 15° \cos 15°\) is:
Answer: B
This is the double-angle formula in reverse.
\[ 2\sin A\cos A = \sin 2A \]
\[ 2\sin 15^{\circ}\cos 15^{\circ} = \sin 30^{\circ} = \frac{1}{2} \]
Recognising the form avoids having to evaluate either 15° value.
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If \(\tan \theta = \frac{a}{b}\), then \(b\cos 2\theta + a\sin 2\theta\) is equal to:
Answer: B
We use the double angle formulas in terms of tan θ:
\(\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}\) and \(\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}\)
Substitute \(\tan \theta = a/b\):
\(\cos 2\theta = \frac{1-a^2/b^2}{1+a^2/b^2} = \frac{b^2-a^2}{b^2+a^2}\)
\(\sin 2\theta = \frac{2(a/b)}{1+a^2/b^2} = \frac{2ab}{a^2+b^2}\)
The expression is \(b(\frac{b^2-a^2}{a^2+b^2}) + a(\frac{2ab}{a^2+b^2}) = \frac{b^3-a^2b+2a^2b}{a^2+b^2} = \frac{b^3+a^2b}{a^2+b^2} = \frac{b(b^2+a^2)}{a^2+b^2} = b\).
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At what angle do the hands of a clock coincide?
Answer: A
When the hands coincide they point in exactly the same direction, so the angle between them is 0°.
They are 180° apart when opposite and 90° apart at a right angle - the other two configurations these questions ask about.
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