(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farm...

(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms, find, correct to: (i) two significant figures, the distance between the two farms, (ii) the nearest degree, the bearing of Manu's farm from Dogbe's.

(b) A ladder 10 m long leaned against a vertical wall xm high. The distance between the wall and the foot of the ladder is 2 m longer than the height of the wall.

Calculate the value of x

Answer Details

(a)

1. (i) To find the distance between the two farms, we can use the law of cosines. Let's call the distance between the two farms "d". Using the law of cosines, we can write:

   d^2 = (5 km)^2 + (3 km)^2 - 2 * (5 km) * (3 km) * cos(200° - 110°)

   Solving for d, we get:

   d = sqrt(58) km = 7.6 km (rounded to two significant figures)

2. (ii) To find the bearing of Manu's farm from Dogbe's, we can use the law of sines. Let's call the angle "theta". Using the law of sines, we can write:

   sin(theta) = (3 km) / d

   Solving for theta, we get:

   theta = arcsin(0.39) = 23° (rounded to the nearest degree)

   So the bearing of Manu's farm from Dogbe's is 23°.

(b)

In this problem, we are trying to find the height of the wall, which we can call "x". We are given that the distance between the wall and the foot of the ladder is 2 m longer than the height of the wall. Let's call this distance "y". So we have:

y = x + 2

We are also given that the ladder is 10 m long. We can use the Pythagorean theorem to find the height of the wall:

x^2 + (10 m)^2 = y^2

Substituting y = x + 2, we get:

x^2 + (10 m)^2 = (x + 2)^2

Expanding and solving for x, we get:

x = 4 m

So the height of the wall is 4 m.