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Heights And Distances | Solved Examples | Trigonometry- Cuemath - Free Printable

Heights And Distances | Solved Examples | Trigonometry- Cuemath

Educational worksheet: Heights And Distances | Solved Examples | Trigonometry- Cuemath. Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Heights And Distances | Solved Examples | Trigonometry- Cuemath
To solve the problem of determining the height of the glacier, we need to analyze the given information and use trigonometry. Let's break it down step by step.

Step 1: Understand the Problem


- We are given a right triangle.
- One of the angles in the triangle is \(45^\circ\).
- The hypotenuse of the triangle is 20 units.
- We need to find the height of the glacier, which corresponds to the vertical side (opposite the \(45^\circ\) angle) of the triangle.

Step 2: Identify the Type of Triangle


Since one of the angles is \(45^\circ\) and it is a right triangle, this is a 45-45-90 triangle. In a 45-45-90 triangle:
- The two legs are congruent.
- The hypotenuse is \(\sqrt{2}\) times the length of each leg.

Step 3: Use the Properties of a 45-45-90 Triangle


Let the length of each leg be \(x\). According to the properties of a 45-45-90 triangle:
\[
\text{Hypotenuse} = x\sqrt{2}
\]
We are given that the hypotenuse is 20 units. Therefore:
\[
x\sqrt{2} = 20
\]

Step 4: Solve for \(x\)


To find \(x\), divide both sides of the equation by \(\sqrt{2}\):
\[
x = \frac{20}{\sqrt{2}}
\]
Rationalize the denominator:
\[
x = \frac{20}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{20\sqrt{2}}{2} = 10\sqrt{2}
\]

Step 5: Determine the Height of the Glacier


In a 45-45-90 triangle, the height (which is one of the legs) is equal to \(x\). Therefore, the height of the glacier is:
\[
10\sqrt{2}
\]

Final Answer:


\[
\boxed{10\sqrt{2}}
\]
Parent Tip: Review the logic above to help your child master the concept of measuring mountains worksheet answers.
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