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Hi!

(i)

The bearing is the angle between North and the segment KL (clockwise)

(ii)

As the segment LM goes straight to South, LM is parallel to North Line. So, the angle KLM is the same of the bearing, or 40 degrees.

(iii)

To find KM we can use cossine theorem.

KM = a = ?

KL = b = 80

LM = c = 120

[tex]a^2=b^2+c^2-2bc\cos\angle KLM\\\\a^2=80^2+120^2-2(80)(120)\cos 40^{\circ}\\\\a^2=6\,400+14\,400-19\,200\cos 40^{\circ}\\\\a^2=20\,800-19\,200\times 0,766044\\\\a^2\approx 6\,091,95\\\\\boxed{a\approx 78,05}[/tex]

I hope I have helped!

(i)

The bearing is the angle between North and the segment KL (clockwise)

(ii)

As the segment LM goes straight to South, LM is parallel to North Line. So, the angle KLM is the same of the bearing, or 40 degrees.

(iii)

To find KM we can use cossine theorem.

KM = a = ?

KL = b = 80

LM = c = 120

[tex]a^2=b^2+c^2-2bc\cos\angle KLM\\\\a^2=80^2+120^2-2(80)(120)\cos 40^{\circ}\\\\a^2=6\,400+14\,400-19\,200\cos 40^{\circ}\\\\a^2=20\,800-19\,200\times 0,766044\\\\a^2\approx 6\,091,95\\\\\boxed{a\approx 78,05}[/tex]

I hope I have helped!

2 posts
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