6. An ostrich cannot fly, but it is able to run fast. Suppose an ostrich runs east for 7.95 s and then runs 161 m south, so that the magnitude of the ostrich's resultant displacement is 226 m. Calculate the magnitude of the ostrich's eastward component.

159 m

Step-by-step explanation:

From the information given:

It was stated that if the ostrich ran towards the east direction in 7.95 s, let say the distance from the starting point is O towards the east side E, let called the distance towards the east side to be OE.

Again, the ostrich then runs in the south direction for 161 m, let the distance be OS.

Also, let the magnitude of the resultant displacement between the east direction to the south direction be ES = 226m.

We are to find, the magnitude of the ostrich's eastward component.

i.e. The distance traveled from the center to the east direction within the time frame of 7.95 s.

Using the Pythagoras rule:

ES² = OE² + OS²

226² = OE² + 161²

226² -  161² = OE²

OE²  =  226² -  161²

OE²  = 51076 - 25921

OE²  =  51076 - 25921

OE²  =  25155

OE =  158.60 m

OE ≅ 159 m

Thus, the magnitude of the ostrich's towards the eastward component. = 159 m.

I am assuming that since the triangle is named BNO, that N is the right angle.  The math works out if N is the right angle, B is the other base angle, and N is the vertex angle.  That means that "side" BO is actually the hypotenuse.  We are looking for the base length, BN.  55 degrees is the reference angle and sinde BN is adjacent to it, so along with that side and the hypotenuse, we will be using the cos ratio.  and 15000cos(55)=x. Solving for x we get x = 8603.6 or 8604 rounded.  First choice above.