Answer:
a. 360.323 m
b. [tex]95.265^{o}[/tex]
Step-by-step explanation:
Distance covered from A to B = speed x time
= 110 x 2.8
= 308 m
Distance covered from B to C = speed x time
= 110 x 1.7
= 187 m
The sum of angles at B = [tex]64^{o}[/tex] + [tex]26^{o}[/tex]
= [tex]90^{o}[/tex]
a. The distance of the plane from it starting point to C can be determined by applying the cosine rule.
[tex]b^{2}[/tex] = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] - 2ac Cos B
Sot hat;
[tex]b^{2}[/tex] = [tex]187^{2}[/tex] + [tex]308^{2}[/tex] - 2(187 x 308) Cos [tex]90^{o}[/tex]
But, Cos [tex]90^{o}[/tex] = 0
So that,
[tex]b^{2}[/tex] = [tex]187^{2}[/tex] + [tex]308^{2}[/tex]
= 34969 + 94864
= 129833
b = [tex]\sqrt{129833}[/tex]
= 360.323
The distance from A to C is 360.323 m.
b. Applying the sine rule;
[tex]\frac{a}{SinA}[/tex] = [tex]\frac{b}{SinB}[/tex]
[tex]\frac{187}{SinA}[/tex] = [tex]\frac{360.323}{Sin90^{o} }[/tex]
[tex]\frac{187}{SinA}[/tex] = [tex]\frac{360.323}{1}[/tex]
Sin A = [tex]\frac{187}{360.323}[/tex]
= 0.5190
⇒ A = [tex]Sin^{-1}[/tex] 0.5190
= [tex]31.265^{o}[/tex]
The bearing of the plane from its original location = [tex]64^{o}[/tex] + [tex]31.265^{o}[/tex]
= [tex]95.265^{o}[/tex]
