From ABC, we have h = x tan 60^ . Since tan 60^ = √(3), we get h = x√(3). From ABD, we have h = (x+20) tan 30^ . Since tan 30^ = {1}{√(3)}, we get h = (x+20) {1}{√(3)}.

Now we equate the two expressions for h from equations (1) and (2). This allows us to solve for x. Multiplying both sides by √(3) eliminates the fraction. Then, we rearrange the terms to isolate x and find its value.

Now we equate the two expressions for h from equations (1) and (2). This allows us to solve for x. Multiplying both sides by √(3) eliminates the fraction. Then, we rearrange the terms to isolate x and find its value.

The angle of elevation of the top of a building from a point on the ground is 30 degrees. On walking 20 m towards the building the angle becomes 60 degrees. Find the height of the building, and draw a diagram.

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Step-by-Step Solution

Step 1: Define variables and set up equations

Let $AB$ be the height of the building, denoted by $h$ . Let $BC$ be the distance from the second observation point to the building, denoted by $x$ . The first observation point is $D$ , so $BD = BC + CD = x + 20$ . We use the tangent function, which relates the opposite side (height) to the adjacent side (distance from the building) in a right-angled triangle.

Step 2: Express height in terms of distance

From $\triangle ABC$ , we have $h = x \tan 60^\circ$ . Since $\tan 60^\circ = \sqrt{3}$ , we get $h = x\sqrt{3}$ . From $\triangle ABD$ , we have $h = (x+20) \tan 30^\circ$ . Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$ , we get $h = (x+20)\frac{1}{\sqrt{3}}$ .

Step 3: Solve for x

Now we equate the two expressions for $h$ from equations (1) and (2). This allows us to solve for $x$ . Multiplying both sides by $\sqrt{3}$ eliminates the fraction. Then, we rearrange the terms to isolate $x$ and find its value.

Step 4: Calculate the height of the building

Substitute the value of $x$ back into equation (1) to find the height $h$ of the building. The height is $10\sqrt{3}$ meters.