Answer to Question #88636 in Mechanics | Relativity for Salim ghali salis

Question

A spelunker os surveying a cave.she follows a passage 180m straight west then 210m in a direction 45 degrees east of south,and then 280m at 30 degrees east of north after a fourth unmeasured displacement,she finds her self back where she started.use a scale drawing to determine the magnitude and direction of the fourth displacement.

Answer 1

From the condition of the problem it is known:

AB = 180 m;

BC = 210 m;

CD = 280 m;

<ABC = 45⁰;

<MCN = 30⁰.

From rectangular triangle CMB:

CM = BC*sin45⁰ = 148.5 m;

From rectangular triangle CMN:

<CNM = 60⁰;

CN = CM/cos30⁰ = 171.5 m;

DN = CD - CN = 108.5 m.

From rectangular triangle NKD:

<DNK = <CNM = 60⁰;

<NDK = 30⁰;

DK = DN*cos30⁰ = 94 m;

NK = DN*sin30⁰ = 54.25 m.

From rectangular triangle CMB:

BM = BC*cos45⁰ = 148.5 m;

AM = AB - BM = 31.5 m.

From rectangular triangle CMN:

MN = CN*sin30⁰ = 85.75 m;

AN = MN - AM = 54.25 m;

AK = NK + AN = 54.25 +54.25 = 108.5 m.

From rectangular triangle AKD:

<ADK = arctan (AK/DK) = 49⁰;

AD = DK/(cos<ADK) = 143.3 m.

So, direction of the fourth displacement is 49⁰ west of south, magnitude of the fourth displacement is 143.3 m.