Mathematics Questions and Answers

\(\bigtriangleup\)PUS is right angled

\(\frac{US}{5}\) = sin60^o

US = 5 x \(\frac{\sqrt{3}}{2}\)

= 2.5\(\sqrt{3}\)

= 4.33cm

The volume of the pipe is equal to the area of the cross section and length.

let outer and inner radii be R and r respectively.

Area of the cross section = (R² - r²)

where R = 6 and r = 6 - 1

= 5cm

Area of the cross section = (6² - 5²)\(\pi\) = (36 - 25)\(\pi\)cm sq

vol. of the pipe = \(\pi\) (R² - r²)L where length (L) = 10

volume = 11\(\pi\) x 10

= 110\(\pi\)cm³

\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i)

from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii)

Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \)

= \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\)

\(2\cos^2 \theta = \frac{7}{8}\)

\(\cos^2 \theta = \frac{7}{2 \times 8}\)

\(\frac{7}{16} = \cos \theta\)

\(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\)

but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\)

\(opp^2 = hyp^2 - adj^2\)

\(opp^2 = 4^2  - (\sqrt{7})^{2}\)

= 16 - 7

opp = \(\sqrt{9}\) = 3

\(\tan \theta = \frac{\text{opp}}{\text{adj}}\)

= \(\frac{3}{\sqrt{7}}\)

\(\frac{3}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)

PR² = PQ² + QR² - 2(QR)(PQ) COS 120^o

PR² = 1² + 2² - 2(1)(2) x - cos 60^o

= 5 - 2(1)(2) x -\(\frac{1}{2}\)

= 5 + 2 = 7

PR = \(\sqrt{7}\)cm

\(\cot \theta = \frac{x}{y}\)

\(\implies \tan \theta = \frac{y}{x}\)

\(opp = y; adj = x\)

Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\)

\(Hyp^{2} = y^{2} + x^{2}\)

\(Hyp = \sqrt{y^{2} + x^{2}}\)

\(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\)

\(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\)

= \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\)