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Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r = 6 \(\theta\) x 2\(\pi\)r = 360 x 6 \(\theta\) = \(\frac{360 \times 6}{2\pi r}\) Area of the sector = \(\frac{\theta}{360}\) x \(\pi\)r2 \(\frac{360 \times 6}{2\pi r}\) x \(\frac{1}{360}\) x \(\pi\)r2 = r = 3 x 7 = 21cm2

\((m + n)^{2} = (m - n)^{2} + x^{2}\) \(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\) \(x^{2} = 4mn\) \(x = \sqrt{4mn} = 2\sqrt{mn}\) 1 + tan2\(\theta\) = sec2\(\theta\) = \(\frac{1}{cos^2\theta}\) \(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\) \(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\) \(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\) = \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)

37.05 x 0.0042 in standard form \(\begin{array}{c|c}No. & log \\\hline 37.05 & 1.5688\\ 0.0042 & 3.6232 \\ \hline & 1.1920\end{array}\) = 0.1556 = 1.556 x 10-1

Cos 60o = \(\frac{PD}{2}\) Cos 60o x 2 = 0.5 x 2 = 1m

Given the above statement, \(y = Q + \frac{P}{x^{2}}\)

Representing from Asia is 150 \(\frac{150}{800}\) x 360 = \(\frac{135}{2}\) = 67\(\frac{1}{2}\)o

Formula for an interior angle of a polygon is (2n - 4) x 90 (2n - 4) x 90 = 144n 180n - 360 = 144n 180n - 144n = 360 36n = 360 n = \(\frac{360}{36}\) = 10

Given: length of the arc AOB = 22cm ; L = 6cm Curved surface area of cone = \(\pi\)rl Length of an arc = \(\frac{\theta}{360}\) x 2 \(\pi\)L = 22cm but length of an arc = circumference of the cone = 2\(\pi\) where r is the radius of the cone circle 2\(\pi\)r = 22, r = \(\frac{22}{2\pi}\)r = 11 x \(\frac{7}{22}\) = \(\frac{7}{2}\) curved surface area = \(\pi\)rl = \(\frac{22}{7}\) x \(\frac{7}{2}\) x 6 = 66 sq.cm

\(\frac{x}{2}\) = sin60o = cos30o x = 2 sino = 2 x \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\) length of the diagonal = 2 x \(\sqrt{2}\) = 2\(\sqrt{3}\)

From \(\bigtriangleup\)QPR, o - (25o + 90o) 180o - 115o = 65o From \(\bigtriangleup\)PSQ, Sin 65o = \(\frac{QPR}{hyp}\) = \(\frac{PS}{5}\) PS = 5 sin 65o From \(\bigtriangleup\)PSR, tan = \(\frac{OPP}{adj}\) = \(\frac{PS}{QS}\) but PS = 5 sin 65o QS tan 25o = PS QS tan 25o = 5 sin 65o QS = \(\frac{5 sin 65^o}{tan 25^o}\) = 5 tan 25o cos 65o

o o o

Area of a //gm PQRS = B X H Area of a //gm PQRS = PQ x TF 50 = 10TF TF = 5cm

In the given figure, QRST is a rectangle with PT parallel to QM and ∠TPQ = 60^\circ. Since PT ∥ QM, and PQ is a transversal, the corresponding angle ∠PQM = 60^\circ, so the remaining angle ∠MQR = 90^\circ − 60^\circ = 30^\circ. Therefore, ∠MUR (the angle formed by the line from M through U to R) is supplementary to this 30^\circ angle, making it 180^\circ − 30^\circ = 150^\circ

\(x \propto \frac{1}{y}\) \(x = \frac{k}{y}\) \(y \propto t^{2}\) \(y = ct^{2}\) k and c are constants. \(x = \frac{k}{ct^{2}}\) Let \(\frac{k}{c} = d\) (a constant) \(x = \frac{d}{t^{2}}\) \(1 = \frac{d}{3^{2}} \implies d = 9\) \(\therefore x = \frac{9}{t^{2}}\) \(x = 9 \div (\frac{1}{3})^{2} \) = \( 9 \div \frac{1}{9} = 9 \times 9 = 81\)

\(\sin x = \cos x\) \(\implies x = 45^\circ\) In radians, \(x = \frac{\pi}{4}\).