Problems about solutions of triangles?

Question

1. A wall extending east and west is 6 feet high. The sun has an altitude of 49degrees32' and is 47degrees20' east of south. Find the width of the shadow of the wall on level ground.

2. A football player stands at a distance c behind the middle of the goal. he sees the angle of elevation of the nearer crossbar to be u and that of the farther one to be v. Show that the distance between the goals is c(tan u cot v - 1).

thanks for your help...i need it urgently...

Answer 1

1.

Let:

O be a point on the base of the wall,

P be the point at the top of the wall vertically above O,

N be a point on the ground with ON perpendicular to the wall,

Q be a point on the ground such that Q is west of N,

u be the measure of angle OQP which is the altitude (49d 32m) of the sun,

v be the bearing (47d 20m) of the sun measured east of south,

h be the height (OP = 6 ft) of the wall,

y be the length of OQ,

x be the length of ON, which is the width of the shadow.

In the right-angled triangle OPQ:

y = h cot(u).

In the right-angled triangle ONQ:

x = y cos(v).

Eliminating y:

x = h cot(u) cos(v)

= 6 cot(49d 32m) cos(47d 20m)

= 3.47 ft.



2.

Let:

O be the position of the observer's feet,

B be the base and A the top of the nearer goal,

D be the base and C the top of the other goal,

x be the required distance between the goals.

In triangle ABO:

h = c tan(u)

In triangle CDO:

h = (x + c)tan(v)

Equating the two values of h:

c tan(u) = (x + c)tan(v)

x tan(v) = c[ tan(u) - tan(v) ]

x = c[ tan(u)cot(v) - 1 ].