Solutions: Introduction to Trigonometry - Test 2

Q1. (a) \(\sin 60^\circ\)

\(\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2}\).
We know \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).

Q2. (b) \(\frac{1}{2}\)

\(\cos^2 30^\circ - \sin^2 30^\circ = (\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\).

Q3. (c) \(0^\circ\)

Substitute \(A = 0^\circ\): LHS = \(\sin 0 = 0\). RHS = \(2 \sin 0 = 0\). LHS = RHS.
For \(A = 30^\circ\), \(\sin 60^\circ = \frac{\sqrt{3}}{2} \neq 2(\frac{1}{2}) = 1\).

Q4. (a) Both A and R are true and R is the correct explanation of A.

\(\tan 60^\circ = \sqrt{3} \approx 1.732 > 1\). Assertion is true.
Value of \(\tan \theta\) increases from 0 to \(\infty\) as \(\theta\) goes from \(0^\circ\) to \(90^\circ\). Reason is true and explains why \(\tan 60^\circ > \tan 45^\circ = 1\).

Q5. Evaluate: \(2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ\).

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

Q6. Find A and B.

\(\sin(A-B) = \frac{1}{2} \Rightarrow A-B = 30^\circ\) ...(1)
\(\cos(A+B) = \frac{1}{2} \Rightarrow A+B = 60^\circ\) ...(2)
Adding (1) and (2): \(2A = 90^\circ \Rightarrow A = 45^\circ\).
Substitute A into (2): \(45^\circ + B = 60^\circ \Rightarrow B = 15^\circ\).

Q7. Find \(x\).

\(\tan 3x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ\)
\(\tan 3x = (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + \frac{1}{2}\)
\(\tan 3x = \frac{1}{2} + \frac{1}{2} = 1\)
\(\tan 3x = \tan 45^\circ \Rightarrow 3x = 45^\circ \Rightarrow x = 15^\circ\).

Q8. Verify \(\cos 60^\circ = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ}\).

LHS = \(\cos 60^\circ = \frac{1}{2}\).
RHS = \(\frac{1 - (1/\sqrt{3})^2}{1 + (1/\sqrt{3})^2} = \frac{1 - 1/3}{1 + 1/3}\)
\(= \frac{2/3}{4/3} = \frac{2}{4} = \frac{1}{2}\).
LHS = RHS. Verified.

Q9. Find BC and AC.

In \(\triangle ABC\), \(\angle B = 90^\circ, \angle C = 30^\circ, AB = 5\) cm.
\(\tan C = \frac{AB}{BC} \Rightarrow \tan 30^\circ = \frac{5}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{5}{BC} \Rightarrow BC = 5\sqrt{3}\) cm.
\(\sin C = \frac{AB}{AC} \Rightarrow \sin 30^\circ = \frac{5}{AC} \Rightarrow \frac{1}{2} = \frac{5}{AC} \Rightarrow AC = 10\) cm.

Q10. Evaluate expression.

Numerator = \(5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ\)
\(= 5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2 - (1)^2 = 5(\frac{1}{4}) + 4(\frac{4}{3}) - 1 = \frac{5}{4} + \frac{16}{3} - 1\)
\(= \frac{15 + 64 - 12}{12} = \frac{67}{12}\).
Denominator = \(\sin^2 30^\circ + \cos^2 30^\circ = 1\) (Identity).
Result = \(\frac{67/12}{1} = \frac{67}{12}\).

Height to reach = \(5 - 1.3 = 3.7\) m.

(i) \(\sin 60^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3.7}{L} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3.7}{L} \Rightarrow L = \frac{7.4}{\sqrt{3}} \approx 4.28\) m.

(ii) \(\cot 60^\circ = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{D}{3.7} \Rightarrow \frac{1}{\sqrt{3}} = \frac{D}{3.7} \Rightarrow D = \frac{3.7}{\sqrt{3}} \approx 2.14\) m.

(iii) \(\sin 30^\circ = \frac{3.7}{L} \Rightarrow \frac{1}{2} = \frac{3.7}{L} \Rightarrow L = 7.4\) m.