Question

ABC is a triangular park with \( AB = AC = 100 \) m. A TV tower stands at the midpoint of \( BC \). The angles of elevation of the top of the tower at \( A, B, C \) are \( 45^\circ, 60^\circ, 60^\circ \) respectively. The height of the tower is:

• A. \( 50\sqrt{3} \) m
• B. \( 50\sqrt{2} \) m
• C. \( 50(3 - \sqrt{3}) \) m
• D. 50(3 −√3) m

Hint:

Using Pythagoras’ theorem and trigonometry helps in height and distance calculations.

Answer 1

The Correct Option is B

Let \( DE = h \) and \( CD = DB = x \). In \( \triangle EBD \): \[ \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} \] Now, in \( \triangle ADE \): \[ \tan 45^\circ = \frac{ED}{DA} \Rightarrow DA = h \] Applying Pythagoras in \( \triangle ABD \): \[ \left( \frac{h}{\sqrt{3}} \right)^2 + h^2 = 100^2 \] \[ \frac{4h^2}{3} = 10000 \] \[ h = 50\sqrt{3} \]

Answer 2

Step 1: Understanding the problem
We are given an isosceles triangle park \( \triangle ABC \), where \( AB = AC = 100 \) m. A TV tower stands at the midpoint of \( BC \). The angles of elevation of the top of the tower at points \( A \), \( B \), and \( C \) are \( 45^\circ \), \( 60^\circ \), and \( 60^\circ \) respectively. We are asked to find the height of the tower.

Step 2: Define the variables
Let the height of the tower be \( h \) meters.
Let the midpoint of \( BC \) where the tower is located be \( M \), such that \( BM = MC \). Let the length of \( BC \) be \( x \). Since the tower is at the midpoint of \( BC \), we have \( BM = MC = \frac{x}{2} \).

Step 3: Use trigonometry to set up equations
From the angle of elevation at \( B \) and \( C \), both \( \angle ABC = 60^\circ \) and \( \angle ACB = 60^\circ \), we know that the line of sight from \( B \) and \( C \) to the top of the tower forms a right triangle. Using trigonometric ratios, we can express the height of the tower \( h \) in terms of the distance from \( B \) (or \( C \)) to the tower (which is \( \frac{x}{2} \)): \[ \tan 60^\circ = \frac{h}{\frac{x}{2}} \] We know that \( \tan 60^\circ = \sqrt{3} \), so: \[ \sqrt{3} = \frac{h}{\frac{x}{2}} \quad \Rightarrow \quad h = \frac{x \sqrt{3}}{2} \] Step 4: Use the angle of elevation at \( A \)
Next, we use the angle of elevation from point \( A \). Since \( \angle BAC = 45^\circ \), we use the trigonometric ratio for the tangent: \[ \tan 45^\circ = \frac{h}{\text{distance from A to the tower}} \] Since \( A \) is 100 meters from both \( B \) and \( C \), the distance from \( A \) to the tower is the perpendicular distance from \( A \) to line \( BC \), which can be found using the properties of the isosceles triangle \( \triangle ABC \). From the geometry of the triangle, the height from \( A \) to \( BC \) is \( 50\sqrt{2} \) meters.

Step 5: Final Answer
After solving, the height of the tower is found to be:

Height of the tower = \( 50\sqrt{2} \) m