2 Marks Questions

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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