width of river

[nickname]

I need help with the following problem:

angela wishes to find the distance from point A to her friend Carmen's house at point C on the other side of the river. She knows the distance from A to Betty's house at B is 540feet. The measurement of angles A and B are 57degrees and 46, respectively. Calculate distane from A to C.

I did:

Angle C= 180-(57+46)= 77degrees

(sinC/AB)=(sinB/AC)--------> AC=(ABsinB/sinC)=(540sin46/sin77)= 398ft, which is the correct answer (is this the correct way of solving this problem?)

Then I have to find the width of the river, assuming that the houses are on the very straight banks of the river? How do I solve this part?

I did (398sin57)/(sin77)= 342.6ft, but the correct answer is 334ft.

 

[mmm4444bot]

… I have to find the width of the river, assuming that the houses are on the very straight banks of the river? How do I solve this part?

I did (398sin57)/(sin77)= 342.6ft, but the correct answer is 334ft.

Not only do we need to assume that all three houses are on straight banks of the river, we also need to assume that the two banks are parallel.

If so, then draw a line segment perpendicular to AB to point C. The length of this line segment is the width of the river.

This line segment divides triangle ABC into two right triangles. Use the right triangle that has AC as its hypotenuse.

Let w = the width of the river

Now use the right-triangle definition of sine:

sin(angle) = opposite/hypotenuse

You know the angle at A.

You know the length of the hypotenuse.

The length of the opposite side is w.

Substitute these into the right-triangle definition of sine, and solve for w.

Thank you for showing your work. Let us know if you need more help with this exercise.