Lines and Angles | FIO

Question 25

Name the different angles in the figure and write their measures.

Solution

We use the protractor to find each angle's measure.

Step 1 — Read ray positions

We place the protractor's center at point O. The baseline of the protractor aligns with line UOP. Ray OP is at 0 degrees on the inner scale. We read each ray's degree mark on the inner scale.

The position of ray OQ is 30 degrees. The position of ray OR is 90 degrees. The position of ray OS is 120 degrees. The position of ray OT is 150 degrees. The position of ray OU is 180 degrees.

Diagram 1

Step 2 — Calculate angle measures

We find the measure of each angle. We subtract the smaller degree from the larger degree.

For angle $\angle POQ$ : $m(\angle POQ) = \text{position of OQ} - \text{position of OP}$ $m(\angle POQ) = 30^\circ - 0^\circ$

$\boxed{m(\angle POQ) = \mathbf{30^\circ}}$

For angle $\angle POR$ : $m(\angle POR) = \text{position of OR} - \text{position of OP}$ $m(\angle POR) = 90^\circ - 0^\circ$

$\boxed{m(\angle POR) = \mathbf{90^\circ}}$

For angle $\angle POS$ : $m(\angle POS) = \text{position of OS} - \text{position of OP}$ $m(\angle POS) = 120^\circ - 0^\circ$

$\boxed{m(\angle POS) = \mathbf{120^\circ}}$

For angle $\angle POT$ : $m(\angle POT) = \text{position of OT} - \text{position of OP}$ $m(\angle POT) = 150^\circ - 0^\circ$

$\boxed{m(\angle POT) = \mathbf{150^\circ}$

For angle $\angle POU$ : $m(\angle POU) = \text{position of OU} - \text{position of OP}$ $m(\angle POU) = 180^\circ - 0^\circ$

$\boxed{m(\angle POU) = \mathbf{180^\circ}}$

For angle $\angle QOR$ : $m(\angle QOR) = \text{position of OR} - \text{position of OQ}$ $m(\angle QOR) = 90^\circ - 30^\circ$

$\boxed{m(\angle QOR) = \mathbf{60^\circ}}$

For angle $\angle QOS$ : $m(\angle QOS) = \text{position of OS} - \text{position of OQ}$ $m(\angle QOS) = 120^\circ - 30^\circ$

$\boxed{m(\angle QOS) = \mathbf{90^\circ}}$

For angle $\angle QOT$ : $m(\angle QOT) = \text{position of OT} - \text{position of OQ}$ $m(\angle QOT) = 150^\circ - 30^\circ$

$\boxed{m(\angle QOT) = \mathbf{120^\circ}$

For angle $\angle QOU$ : $m(\angle QOU) = \text{position of OU} - \text{position of OQ}$ $m(\angle QOU) = 180^\circ - 30^\circ$

$\boxed{m(\angle QOU) = \mathbf{150^\circ}$

For angle $\angle ROS$ : $m(\angle ROS) = \text{position of OS} - \text{position of OR}$ $m(\angle ROS) = 120^\circ - 90^\circ$

$\boxed{m(\angle ROS) = \mathbf{30^\circ}}$

For angle $\angle ROT$ : $m(\angle ROT) = \text{position of OT} - \text{position of OR}$ $m(\angle ROT) = 150^\circ - 90^\circ$

$\boxed{m(\angle ROT) = \mathbf{60^\circ}$

For angle $\angle ROU$ : $m(\angle ROU) = \text{position of OU} - \text{position of OR}$ $m(\angle ROU) = 180^\circ - 90^\circ$

$\boxed{m(\angle ROU) = \mathbf{90^\circ}$

For angle $\angle SOT$ : $m(\angle SOT) = \text{position of OT} - \text{position of OS}$ $m(\angle SOT) = 150^\circ - 120^\circ$

$\boxed{m(\angle SOT) = \mathbf{30^\circ}$

For angle $\angle SOU$ : $m(\angle SOU) = \text{position of OU} - \text{position of OS}$ $m(\angle SOU) = 180^\circ - 120^\circ$

$\boxed{m(\angle SOU) = \mathbf{60^\circ}$

For angle $\angle TOU$ : $m(\angle TOU) = \text{position of OU} - \text{position of OT}$ $m(\angle TOU) = 180^\circ - 150^\circ$

$\boxed{m(\angle TOU) = \mathbf{30^\circ}$

Answer

(i) $\angle POQ = \mathbf{30^\circ}$ (ii) $\angle POR = \mathbf{90^\circ}$ (iii) $\angle POS = \mathbf{120^\circ}$ (iv) $\angle POT = \mathbf{150^\circ}$ (v) $\angle POU = \mathbf{180^\circ}$ (vi) $\angle QOR = \mathbf{60^\circ}$ (vii) $\angle QOS = \mathbf{90^\circ}$ (viii) $\angle QOT = \mathbf{120^\circ}$ (ix) $\angle QOU = \mathbf{150^\circ}$ (x) $\angle ROS = \mathbf{30^\circ}$ (xi) $\angle ROT = \mathbf{60^\circ}$ (xii) $\angle ROU = \mathbf{90^\circ}$ (xiii) $\angle SOT = \mathbf{30^\circ}$ (xiv) $\angle SOU = \mathbf{60^\circ}$ (xv) $\angle TOU = \mathbf{30^\circ}$

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Q12

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Q18

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Q41

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Q42

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Q44

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