Using Bearings and Distances to Find Angles

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Question 1 of 2

1. Question
Find the value of $∠ P Q R$ $angle PQR$ using the bearing of $Q$ $Q$ from $P$ $P$ .
- $∠ P Q R =$ $/_PQR=$ (146) $°$ $°$
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First, we can add a vertical line that goes through point $Q$ $Q$ .

add a vertical line through point Q

This will indicate that part of $∠ P Q R$ $angle PQR$ is a right angle, which is $90 °$ $90°$ .

Next, notice that the two vertical lines are parallel.

From there, we can see that a part of $∠ P Q R$ $angle PQR$ is an alternate angle to $∠ N P Q$ $angle NPQ$ .

Since alternate angles are equal, the missing angle also measures $∙ 56 °$ $• 56°$ .

Finally, we can add $∙ 56 °$ $• 56°$ to the right angle to find the value of $∠ P Q R$ $angle PQR$ .

$∠ P Q R$ $/_PQR$ $=$ $=$ $90 ° + 56 °$ $90°+56°$

$=$ $=$ $146 °$ $146°$

$∠ P Q R = 146 °$ $/_ PQR=146°$
Question 2 of 2

2. Question
Find the value of interior angles $X$ $X$ , $Y$ $Y$ , and $Z$ $Z$ using the given bearings.
- $∠ X =$ $angleX=$ (49) $°$ $°$
  
  $∠ Y =$ $angleY=$ (40) $°$ $°$
  
  $∠ Z =$ $angleZ=$ (91) $°$ $°$
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Complementary angles are when two angles have a sum of $90 ° .$ $90°.$ Typically, these angles form a right angle.

The sum of the interior angles in a triangle is $180 °$ $180°$ .

First, notice that $∠ X$ $angleX$ and the given bearing of $B$ $B$ from $A$ $A$ are complementary angles.

Since complementary angles are equal to $90 °$ $90°$ , we can simply subtract the bearing of $B$ $B$ from $A$ $A$ to find the measure of $∠ X$ $angleX$

$∠ X$ $angleX$ $=$ $=$ $90 ° - 41 °$ $90°-41°$

$=$ $=$ $49 °$ $49°$

Next, notice that $∠ Y$ $angleY$ and the given bearing of $B$ $B$ from $C$ $C$ complementary angles.

Since complementary angles are equal to $90 °$ $90°$ , we can simply subtract the bearing of $B$ $B$ from $C$ $C$ to find the measure of $∠ Y$ $angleY$

$∠ Y$ $angleY$ $=$ $=$ $90 ° - 50 °$ $90°-50°$

$=$ $=$ $40 °$ $40°$

Finally, to find the measure of $∠ Z$ $angleZ$ , subtract the sum of $∠ X$ $angleX$ and $∠ Y$ $angleY$ from $180 °$ $180°$ , which is the sum of the interior angles of a triangle.

$∠ Z$ $angleZ$ $=$ $=$ $180 ° - (49 ° + 40 °)$ $180°-(49°+40°)$

$=$ $=$ $180 ° - 89 °$ $180°-89°$

$=$ $=$ $91 °$ $91°$

$∠ X = 49 °$ $angleX=49°$

$∠ Y = 40 °$ $angleY=40°$

$∠ Z = 91 °$ $angleZ=91°$

Using Bearings and Distances to Find Angles

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2. Question

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$∠ P Q R$ $/_PQR$	$=$ $=$	$90 ° + 56 °$ $90°+56°$
	$=$ $=$	$146 °$ $146°$

$∠ X$ $angleX$	$=$ $=$	$90 ° - 41 °$ $90°-41°$
	$=$ $=$	$49 °$ $49°$

$∠ Y$ $angleY$	$=$ $=$	$90 ° - 50 °$ $90°-50°$
	$=$ $=$	$40 °$ $40°$

$∠ Z$ $angleZ$	$=$ $=$	$180 ° - (49 ° + 40 °)$ $180°-(49°+40°)$
	$=$ $=$	$180 ° - 89 °$ $180°-89°$
	$=$ $=$	$91 °$ $91°$