A Water Tank Of Height 10 Ft Is Fixed On The Roof Of A Building. The Angles Subtended By The Building And The Tank At A Point On The Ground Are $45^{\circ}$ And $60^{\circ}$, Respectively. Find The Height Of The Building.Answer:

Introduction

In this article, we will delve into a problem involving the geometry of a building and a water tank. The problem states that a water tank of height 10 ft is fixed on the roof of a building, and the angles subtended by the building and the tank at a point on the ground are $45^{\circ}$ and $60^{\circ}$, respectively. Our goal is to find the height of the building.

Understanding the Problem

To approach this problem, we need to understand the given information and the relationships between the different components involved. We are given a water tank with a height of 10 ft, and it is fixed on the roof of a building. The angles subtended by the building and the tank at a point on the ground are $45^{\circ}$ and $60^{\circ}$, respectively. This information provides us with a starting point to work with.

Using Trigonometry to Solve the Problem

One way to approach this problem is by using trigonometry. We can start by drawing a diagram that represents the situation. Let's assume that the point on the ground is at a distance of x from the building. We can then draw a line from the point on the ground to the top of the building and another line from the point on the ground to the top of the tank.

Using the given angles, we can form two right-angled triangles. The first triangle has an angle of $45^{\circ}$ and the second triangle has an angle of $60^{\circ}$. We can use the trigonometric ratios to relate the angles, the height of the building, and the distance x.

Applying the Trigonometric Ratios

Let's start by applying the trigonometric ratios to the first triangle. We can use the tangent function to relate the angle, the height of the building, and the distance x. The tangent function is defined as the ratio of the opposite side to the adjacent side.

For the first triangle, we have:

$$\tan(45^{\circ}) = \frac{10}{x}$$

We know that $\tan(45^{\circ}) = 1$, so we can simplify the equation to:

$$1 = \frac{10}{x}$$

Solving for x

To solve for x, we can multiply both sides of the equation by x, which gives us:

$$x = 10$$

However, this is not the correct value for x. We need to consider the second triangle and apply the trigonometric ratios to find the correct value for x.

Applying the Trigonometric Ratios to the Second Triangle

For the second triangle, we have:

$$\tan(60^{\circ}) = \frac{10}{x + h}$$

We know that $\tan(60^{\circ}) = \sqrt{3}$, so we can simplify the equation to:

$$\sqrt{3} = \frac{10}{x + h}$$

Solving for x and h

To solve for x and h, we can multiply both sides of the equation by (x + h), which gives us:

$$(x + h)\sqrt{3} = 10$$

We can then square both sides of the equation to get:

$$(x + h)^2 = \frac{100}{3}$$

Expanding and Simplifying the Equation

We can expand and simplify the equation to get:

$$x^2 + 2hx + h^2 = \frac{100}{3}$$

Using the Pythagorean Theorem

We can use the Pythagorean theorem to relate the height of the building, the distance x, and the distance h. The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

For the second triangle, we have:

$$h^2 + x^2 = (x + h)^2$$

We can simplify the equation to get:

$$h^2 + x^2 = x^2 + 2hx + h^2$$

Cancelling Out Terms

We can cancel out the terms h^2 and x^2 on both sides of the equation, which gives us:

$$0 = 2hx$$

Solving for h

To solve for h, we can divide both sides of the equation by 2x, which gives us:

$$h = \frac{0}{2x}$$

However, this is not a valid solution. We need to reconsider our approach and find a different way to solve for h.

Reconsidering the Approach

Let's reconsider our approach and use a different method to solve for h. We can start by using the fact that the tangent function is equal to the ratio of the opposite side to the adjacent side.

For the second triangle, we have:

$$\tan(60^{\circ}) = \frac{10}{x + h}$$

We know that $\tan(60^{\circ}) = \sqrt{3}$, so we can simplify the equation to:

$$\sqrt{3} = \frac{10}{x + h}$$

Multiplying Both Sides by (x + h)

We can multiply both sides of the equation by (x + h), which gives us:

$$(x + h)\sqrt{3} = 10$$

Squaring Both Sides

We can square both sides of the equation to get:

$$(x + h)^2 = \frac{100}{3}$$

Expanding and Simplifying the Equation

We can expand and simplify the equation to get:

$$x^2 + 2hx + h^2 = \frac{100}{3}$$

Using the Pythagorean Theorem

We can use the Pythagorean theorem to relate the height of the building, the distance x, and the distance h. The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

For the second triangle, we have:

$$h^2 + x^2 = (x + h)^2$$

We can simplify the equation to get:

$$h^2 + x^2 = x^2 + 2hx + h^2$$

Cancelling Out Terms

We can cancel out the terms h^2 and x^2 on both sides of the equation, which gives us:

$$0 = 2hx$$

Solving for h

To solve for h, we can divide both sides of the equation by 2x, which gives us:

$$h = \frac{0}{2x}$$

However, this is not a valid solution. We need to reconsider our approach and find a different way to solve for h.

Using Similar Triangles

Let's use similar triangles to solve for h. We can start by drawing a diagram that represents the situation. Let's assume that the point on the ground is at a distance of x from the building. We can then draw a line from the point on the ground to the top of the building and another line from the point on the ground to the top of the tank.

Using the given angles, we can form two similar triangles. The first triangle has an angle of $45^{\circ}$ and the second triangle has an angle of $60^{\circ}$. We can use the properties of similar triangles to relate the angles, the height of the building, and the distance x.

Applying the Properties of Similar Triangles

For the two similar triangles, we have:

$$\frac{h}{x} = \frac{10}{x + h}$$

We can simplify the equation to get:

$$h(x + h) = 10x$$

Expanding and Simplifying the Equation

We can expand and simplify the equation to get:

$$hx + h^2 = 10x$$

Rearranging the Equation

We can rearrange the equation to get:

$$h^2 + hx - 10x = 0$$

Factoring the Equation

We can factor the equation to get:

$$(h + 10)(h - x) = 0$$

Solving for h

To solve for h, we can set each factor equal to zero and solve for h. We get:

$$h + 10 = 0 \Rightarrow h = -10$$

$$h - x = 0 \Rightarrow h = x$$

However, the second solution is not valid because the height of the building cannot be equal to the distance x.

Using the Quadratic Formula

We can use the quadratic formula to solve for h. The quadratic formula is given by:

$$h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, we have:

$$a = 1, b = 1, c = -10$$

We can plug these values into the quadratic formula to get:

$$h = \frac{-1 \pm \sqrt{1 + 40}}{2}$$

Simplifying the Equation

We can simplify the equation to get:

$$h = \frac{-1 \pm \sqrt{41}}{2}$$

Solving for h

To solve for h, we can take the positive square root, which gives us:

$$h = \frac{-1 + \sqrt{41}}{2}$$

Calculating the Value of h

We can calculate the value of h by plugging in the values into a calculator. We get:

$$h \approx 5.52$$

Conclusion

In this article, we used trigonometry and similar

Introduction

In our previous article, we explored a problem involving the geometry of a building and a water tank. The problem states that a water tank of height 10 ft is fixed on the roof of a building, and the angles subtended by the building and the tank at a point on the ground are $45^{\circ}$ and $60^{\circ}$, respectively. Our goal was to find the height of the building.

In this article, we will provide a Q&A section to help clarify any doubts and provide additional insights into the problem.

Q: What is the main concept used to solve this problem?

A: The main concept used to solve this problem is trigonometry, specifically the tangent function.

Q: Why is the tangent function used in this problem?

A: The tangent function is used in this problem because it relates the angle, the height of the building, and the distance x.

Q: What is the significance of the angles $45^{\circ}$ and $60^{\circ}$ in this problem?

A: The angles $45^{\circ}$ and $60^{\circ}$ are significant in this problem because they are used to form two right-angled triangles. The tangent function is then used to relate the angles, the height of the building, and the distance x.

Q: How is the height of the building related to the distance x?

A: The height of the building is related to the distance x through the tangent function. Specifically, the tangent function is used to relate the angle, the height of the building, and the distance x.

Q: What is the relationship between the two similar triangles in this problem?

A: The two similar triangles in this problem are related through the properties of similar triangles. Specifically, the ratio of the corresponding sides of the two triangles is equal.

Q: How is the height of the building calculated in this problem?

A: The height of the building is calculated in this problem using the quadratic formula. The quadratic formula is used to solve for the height of the building, which is given by the equation $h = \frac{-1 \pm \sqrt{41}}{2}$.

Q: What is the final answer to this problem?

A: The final answer to this problem is $h \approx 5.52$.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

• Not using the correct trigonometric function (tangent) to relate the angle, the height of the building, and the distance x.
• Not considering the properties of similar triangles.
• Not using the quadratic formula to solve for the height of the building.

Q: What are some real-world applications of this problem?

A: Some real-world applications of this problem include:

• Architecture: The problem can be used to calculate the height of a building given the angle of elevation and the distance from the building.
• Engineering: The problem can be used to calculate the height of a structure given the angle of elevation and the distance from the structure.
• Surveying: The problem can be used to calculate the height of a landmark given the angle of elevation and the distance from the landmark.

Q: What are some tips for solving this problem?

A: Some tips for solving this problem include:

• Make sure to use the correct trigonometric function (tangent) to relate the angle, the height of the building, and the distance x.
• Consider the properties of similar triangles.
• Use the quadratic formula to solve for the height of the building.

Conclusion

In this article, we provided a Q&A section to help clarify any doubts and provide additional insights into the problem. We hope that this article has been helpful in understanding the problem and its solution.