A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60∘. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30∘ (see Fig.). Find the height of the tower and the width of the canal.

To solve this problem, we can use the properties of right triangles and the definitions of the trigonometric functions sine, cosine, and tangent.

Step 1: Let's denote the height of the tower as h and the width of the canal as d.

Step 2: From the problem, we know that the angle of elevation of the top of the tower from the point directly opposite the tower is 60 degrees. This forms a right triangle with the tower and the canal. So, we can use the tangent of the angle to find the height of the tower.

The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side. So, we have:

tan(60) = h/d

Step 3: From the problem, we also know that the angle of elevation of the top of the tower from another point 20 m away is 30 degrees. This also forms a right triangle with the tower and the line from the point to the foot of the tower. So, we can use the tangent of this angle to find the height of the tower.

tan(30) = h/(d+20)

Step 4: Now we have a system of two equations with two unknowns (h and d). We can solve this system to find the values of h and d.

From tan(60) = h/d, we get h = d * sqrt(3)

And from tan(30) = h/(d+20), we get h = (d+20) * 1/ sqrt(3)

Setting these two equal gives d * sqrt(3) = (d+20) * 1/ sqrt(3)

Solving this equation gives d = 20 m

Substituting d = 20 m into the equation h = d * sqrt(3) gives h = 20 * sqrt(3) m

So, the height of the tower is 20 * sqrt(3) m and the width of the canal is 20 m.