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Cos \(\theta\) = \(\frac{12}{13}\) x\(^2\) + 12\(^2\) = 13\(^2\) x\(^2\) = 169- 144 = 25 x = \(\sqrt{25}\) = 5 Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\) If cot\(^2\)\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\) = 1 + \(\frac{1}{\frac{(5)^2}{12^2}}\) = 1 + \(\frac{1}{\frac{25}{144}}\) = 1 + \(\frac{144}{25}\) = \(\frac{25 + 144}{25}\) = \(\frac{169}{25}\)

yzx + 105o + 30o = 180o yzx = 180 - 155 = 45o \(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\) yz = \(\frac{8 \sin 30}{sin 45}\) = 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\) = 4 \(\div\) \(\frac{1}{\sqrt{2}}\) = 4\(\sqrt{2}\)cm

\(\frac{x}{4} = \frac{6 + x}{6}\) 6x = 4(6 + x) = 24 + 4x x = 12cm CSA = \(\pi RL - \pi rl\) = \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\) = 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\) = 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\) = 20\(\pi \sqrt{10}\)cm2

loga3 + loga5 - loga6 log(a3 * a5/a6) = loga64 a2 = 64 a = 8

No Log 5.34 0.7275 + 67.4 1.8287 2.5562 - 2.7 0.4314 Antilog = 133.2 2.1248 \(\therefore \frac{5.34 \times 67.4}{2.7} = 133.2\)

3\(^y\) = 243; 3\(^y\) = 3\(^5\) y = 5

\(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) = \(\frac{3^{-1}}{3^2}\) = \(\frac{1}{27}\) In full: To simplify the expression : \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) We start by rewriting the bases in terms of powers of 3. We know that: \(9 = 3^2 \quad \text{and} \quad 27 = 3^3\) Thus, we can express \(9^{-\frac{1}{2}}\) as follows: \(9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{-1}\) Next, we rewrite \(27^{\frac{2}{3}}\): \(27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{2}\) Substitute back into the expression: Now we substitute these values back into the original expression: \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}} = \frac{3^{-1}}{3^{2}}\) This simplifies to: \(\frac{1/3}{9} = \frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}\) Thus, the simplified form of \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) is \(\frac{1}{27}\).

\(\log_{a} x = p\) \(x = a^p\)

log p = 2log x + 3log q log p = logx 2 + logq 3 logp = log x 2q 3 p = x 2q 3

\(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\) = \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\) = \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\) Rationalizing, \((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\) = \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\) = \(\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\)

\(\log_{3} p + 3\log_{3} q = 3\) \(\log_{3} p + \log_{3} q^{3} = 3\) \(\implies \log_{3} (pq^{3}) = 3\) \(pq^{3} = 3^{3} = 27\) \(\therefore p = \frac{27}{q^{3}}\) = \((\frac{3}{q})^{3}\)

\(9^{y} - 4 \times 3^{y} + 3 = 0\) \(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\) \((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\) Let \(3^{y}\) be r. Then, \(r^{2} - 4r + 3 = 0\) Solving the equation, \(r^{2} - 3r - r + 3 = 0\) \(r(r - 3) - 1(r - 3) = 0\) \((r - 3)(r - 1) = 0\) \(\therefore \text{r = 3 or 1}\) Recall, \(3^{y} = r\) \(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\) \(\implies \text{y = 1 or 0}\)

\(\frac{324 - 4x^2}{2x + 18}\) = \(\frac{18^2 - (2x)^2}{2x + 18}\) = \(\frac{(18 - 2x)(18 + 2x)}{(2x + 18)}\) 18 - 2x = 2(9 - x) or -2(x - 9)

\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\) Let \(x^{-1}\) = a and \(y^{-1}\)= b 2a - 3b = 4 .......(i) 4a + b = 1 .........(ii) (i) x 3 = 12a + 3b = 3........(iii) 2a - 3b = 4 ...........(i) (i) + (iii) = 14a = 7 ∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\) From (i), 3b = 2a - 4 3b = 1 - 4 3b = -3 ∴ b = -1 From substituting, \(2^{-1} = x^{-1}\) ∴ x = 2 \(y^{-1} = -1, y = -1\)

Y \(\to\) 30^\circ X \(\to\) 120^\circ Z \(\to\) 30^\circ \(\frac{3}{\text{sin 30}}\) = \(\frac{YZ}{\text{sin 120}}\) YZ = \(\frac{3\times \sin 120}{\sin 30}\) \(3 \times \frac{\sqrt{3}}{2} \times 2 = 3\sqrt{3}\)