Question

Every day going to school, June has to go by boat (point A) over a 217m-wide river to point B (the other bank), and then from B walk to school at point D with distance AD ​​= 170m. In fact, due to the flowing water, the boat was pushed by the water at an angle of 50 degrees and brought her to point C (the other shore). From C, June walk to school. Calculate the distance June traveled from A to D. (result rounded to the third decimal place)​

Answer 1

• 647m

Step-by-step explanation:

The distance traveled by June, from A to D, is the sum of the distance from A to C plus the distance from C to D:

• Travel from A to D = Distance AC + Distance DC.

1) Find the distance AC

• Triangle ABC is a right triangle, from which you have to calculate the hypotenuse (AC), knowing angle A (50º), and its adjacent leg (217 m).

• cos (50º) = 217m / AC

• AC = 217m / cos (50º) = 337.59m

2) Find the distance DC

First, you need to calculate the side BC which is a common side of both triangles ABC and DBC.

You can use Pythagora's or a trigonometric function (tangent or sine).

• tan (50º) = BC / 217 m ⇒ BC = 217 m × tan (50º) = 258.61 m

Now you can use Pythagora's theorem to find the distance DC, which is the hypotenuse of the triangle DBC:

• DC² = (170m)² + (258.62)² = 95,779.41m²

• DC = 309.48m

3) Find the total distance traveled from A to D

• Travel from A to D = Distance AC + Distance DC = 337.59m + 309.48m = 647.07 m ≈ 647m.