Three of the four towns are on the vertices of the triangle ΔCBD, through
which the bearing can calculated.
<h3>Response:</h3>
- The bearing of D from B is approximately <u>209.05°</u>
<h3>Method by which the bearing is found;</h3>
From the given information, we have;
AC = AB = 25 km
∠BAC = 90° (definition of angle between north and east)
ΔABC = An isosceles right triangle (definition)
∠ACD = ∠ABC = 45° (base angles of an isosceles right triangle)
![tan(\angle ABD) = \dfrac{45}{25}](https://tex.z-dn.net/?f=tan%28%5Cangle%20ABD%29%20%3D%20%5Cdfrac%7B45%7D%7B25%7D)
The bearing of <em>D</em> from <em>B</em> is the angle measured from the north of <em>B</em> to the
direction of <em>D.</em>
<em />
Therefore;
- The bearing of D from B ≈ 90° + (180° - 60.945°) = <u>209.05°</u>
Learn more about bearings in mathematics here:
brainly.com/question/10710413
I hope this helps you
perpendicular lines slopes multiplication -1
M1×M2= -1
1/2×M2= -1
M2= -2
<span>3^2 + 4^2 = 5^2
answer is </span>A) 3^2 + 4^2 = 5^2
These three angles are on a line so must total 180°.
So 180-(52+58)=
180-110=70
So angle q is 70°.