The diagram shows a 5 cm x 5 cm x 5 cm cube.

Mathematics

1 answer:

Nitella [24]3 years ago

7 0

Answer:

8.7 cm

Step-by-step explanation:

The question is a 2-two-step Pythagoras theorem. (c^2 = a^2 + b^2)

Consider as such, If I were to draw a diagonal line along the base of the cube what is the length of the diagonal line. To find out that we use the theorem. We can substitute a for 5 and b for 5 as well. So

a^2 +b^2 = c^2

5^2 + 5^2 = c^2

25 + 25 = c^2

√50 = c

Then since the line side of the cube is on a 3d angle we need to do the same process again but now using the imaginary diagonal line that we just calculated and the height (5).

a^2 +b^2 = c^2

√50^2 + 5^2 = c^2

50 + 25 = c^2

√75 = c

c = 8.6602...

<em>when rounded to 1 d.p.</em>

c = 8.7

Line AB is 8.7 cm long.

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The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further from the lighthouse. The new bearing is 25°

djverab [1.8K]

Answer:

The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.

Step-by-step explanation:

The question is incomplete. The complete question should be

The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?

Given the initial bearing of a lighthouse from the ship is N 37° E. So, $\angle ABN$ is 37°. We can see from the diagram that $\angle ABC$ would be $180-37=$ 143°.