A point p is on the bearing of 335 from a point Q find the bearing of Q from P
1 answer:
Answer:
The bearing of Q from P is 155°.
Step-by-step explanation:
In the attached photo,
Fig 1. gives details about the question and fig 2. gives the answer to the question.
From fig 1:
b° + 335° = 360° (sum of angle at a point)
b° + 335° = 360°
Collect like terms
b° = 360° – 335°
b° = 25°
a° + b° = 180° (sum of angle in a straight line)
a° + b° = 180°
b° = 25°
a° + 25° = 180°
Collect like terms
a° = 180° – 25°
a° = 155°
a° gives the bearing of Q from P.
Therefore, the bearing of Q from P is 155°.
Fig 2 on the attached photo shows the the bearing of Q from P.
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<em>r = -</em><em />
<em>Step-by-step explanation:</em>
<em>Rewrite the equation as </em><em> = m</em>
<em>Remove the radical on the left side of the equation by squaring both sides of the equation.</em>
<em>(</em><em> = m^2</em>
<em>Then, you simplify each of the equation. </em>
<em>Rewrite: (</em><em> as </em>
<em />
<em>Remove any parentheses if needed.</em>
<em>Solve for r. </em>
<em>Multiply each term by r and simplify."</em>
<em>Multiply both sides of the equation by 5.</em>
<em>6a+r= m^2r⋅(5)</em>
<em>Remove parentheses.</em>
<em>Move 5 to the left of (m ^2) r </em>
<em>6a+r=5m^2)r</em>
<em>Subtract 5m^2)r from both sides of the equation.</em>
<em>6a+r-5m^2)r=0</em>
<em>Subtract 6a from both sides of the equation.</em>
<em>r-5m^2)r=-6a</em>
<em>Factor r out of r-5m^2)r </em>
<em>r(1-5m^2)=-6a</em>
Divide each term by 1-5m^2 and simplify.
r = -
There you go, hope this helps!