Question

Write your question here (Keep it simple and clear to get the best answer)A surveyor leaves her base camp and drives 42km on a bearing of 032degree, she then drives 28km on a bearing of 154degree

Answer 1

Answer:

The surveyor is 36.1 km from the base camp and the base camp is a bearing of 253.4° away.

Step-by-step explanation:

Complete Question

A surveyor leaves her base camp and drives 42km on a bearing of 032°. She then drives 28km on a bearing of 154°. How far is she from her base camp and what is her bearing from it?

Solution

The diagram of the surveyor's movement is attached to this solution of the problem

From the attached image, the complete travel of the surveyor forms a triangle,

Naming the distance from her base camp y

Using Cosine rule

y² = 42² + 28² - (2×42×28×cos 58°)

y² = 2,548 - 1,246.37 = 1,301.63

y = √1,301.63 = 36.1 km

To obtain the surveyor's bearing from her base camp now, we use sine rule

[(Sin 58°)/y] = [(Sin a)/42]

Sin a = (42 × sin 58°)/36.1

a = sin⁻¹ (0.9866)

a = 80.6°

Bearing of the surveyor from the base camp = 270° - (80.6° - 64°) = 253.4°

Hope this Helps!!!

Answer 2

A surveyor leaves her base camp and drives 42 km on a bearing of 032°, she then drives 28 km on a bearing of 154°. How far is she then from her base camp and what is her bearing from it?

Answer:

Thus; she is 31.3 km far distant from the base camp

the surveyor is on the bearing of  64° East North from the base camp

Step-by-step explanation:

If interior part of  angle A = 58° and exterior part of angle A = 32° Then the interior part of angle B = 32° in the east west direction.   (alternate angles)

Now the total angle covered by angle B = ( 32° + (180° -154°)

= (32° + 26°)

=58°

To determine the angle C which is the bearing of the surveyor from her base camp; we have :

58° + 58° + x = 180°   (sum of angles in a triangle)

116° + x = 180°

x = 180° - 116°

x = 64°

Thus; the surveyor is on the bearing of  64° East North from the base camp

Using Pythagoras rule to determine how far she is from the base camp;

we have:

a²  +   b²  = c²

28² + b² = 42²

b² = 42² -   28²

b² = 1764-784

b² = 980

b = $\sqrt{980}$

b = 31.3 km

Thus; she is 31.3 km far distant from the base camp