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Measuremenet Of Angle · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 4 questions.

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Question 12 Marks
Find the magnitude, in radians and degrees, of the interior angle of a regular. Octagon.
Answer
General formula for in angles of polygon with n side $=\Big(\frac{2\text{n}-4}{\text{n}}\Big)\times90^{\circ}$ Pentagon has 5 sides, $\text{n}=8$ $\therefore$ Each angle $=\frac{2\times8-4}{8}\times\frac{\pi}{2}$ $=\Big(\frac{3\pi}{4}\Big)^{\text{c}}$ $\therefore 135^{\circ},\Big(\frac{3\pi}{4}\Big)^{\text{c}}$
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Question 22 Marks
Find the magnitude, in radians and degrees, of the interior angle of a regular. heptagon.
Answer
General formula for in angles of polygon with n side $=\Big(\frac{2\text{n}-4}{\text{n}}\Big)\times90^{\circ}$ Pentagon has 5 sides, $\text{n}=7$ $\therefore$ Each angle $=\frac{2\times7-4}{8}\times90^{\circ}$ $=\frac{10}{7}\times90^{\circ}$ $=128^{\circ}34'17''$ Again, Each angle $=\frac{2\times7-4}{8}\times\frac{\pi}{2}$ $=\frac{10}{7}\times\frac{\pi}{2}$ $=\Big(\frac{5\pi}{7}\Big)^{\text{c}}$ $\therefore 128^{\circ}34'17'',\Big(\frac{5\pi}{7}\Big)^{\text{c}}$
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Question 32 Marks
The difference between the two acute angkes of a right angles triangle is $\frac{2\pi}{5}$ redians. Express the angles in degrees.
Answer
Let $\theta_{1}$ and $\theta_{2}$ be two acute angles of a right angles triangle. $\therefore$ Difference of acute angles. $\theta_{1}-\theta_{2}=\frac{2\pi}{5}\ \text{radians}$ $\therefore$ In a right angled triangle, $\theta_{1}+\theta_{2}=\frac{\pi}{2}$ $\theta_{1}+\theta_{2}=\frac{2\pi}{5}$ $\theta_{1}+\theta_{2}=\frac{\pi}{2}$ On solving $2\theta_{1}=\frac{2\pi}{5}+\frac{\pi}{2}$ $\theta_{1}=\frac{9\pi}{20}$ From quation (ii) $\theta_{2}=\frac{\pi}{20}$ So angles in degrees, $\theta_{1}=\frac{9\pi}{20}\times\frac{180}{\pi}=81^{\circ}$ $\theta_{2}=\frac{\pi}{20}\times\frac{180}{\pi}=9^{\circ}$
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Question 42 Marks
Find the radian measure corresponding to the following degree measures: -47°30'
Answer
We have, $180^{\circ}=\pi^{\text{c}}$ $\therefore 1^{\circ}=\Big(\frac{\pi}{180}\Big)^{\text{c}}$ Now, $-47^{\circ}30^{'}=-47^{\circ}\Big(\frac{30}{60}\Big)^{\circ}$ $=\Big(-47\frac{1}{2}\Big)^{\circ}$ $=\Big(\frac{-95}{2}\Big)^{\circ}$ $=\Big(\frac{-95}{2}\times\frac{\pi}{180}\Big)^{\text{c}}$ $=\frac{-19\pi}{72}$
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