Question

3. A ship sails 35 km on a bearing of 042º.

Answer 1

Answer:

3(a). The ship travelled in north is 26.0 km.

(b). The ship travelled in east is 23.4 km.

4(a). The ship travelled in south is -88.61 km.

(b). The ship travelled in west is -179.29 km.

Step-by-step explanation:

Given that,

(3). Distance = 35 km

Angle = 42°

Let distance in north is y km.

We need to calculate the distance

Using vertical component

$y = d\cos\theta$

Put the value into the formula

$y = 35\cos42$

$y=26.0\ km$

Let distance in east is x  km

We need to calculate the distance

Using horizontal component

$x =d\sin\theta$

Put the value into the formula

$x = 35\sin42$

$x=23.4\ km$

(4). A ship sails 200 km on a bearing of 243.7°

Let distance in south is y km.

We need to calculate the distance

Using vertical component

$y = d\cos\theta$

Put the value into the formula

$y = 200\cos243.7$

$y=-88.61\ km$

Let distance in west is x  km

We need to calculate the distance

Using horizontal component

$x =d\sin\theta$

Put the value into the formula

$x = 200\sin243.7$

$x=-179.29\ km$

Hence, 3(a). The ship travelled in north is 26.0 km.

(b). The ship travelled in east is 23.4 km.

4(a). The ship travelled in south is -88.61 km.

(b). The ship travelled in west is -179.29 km.