/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Find the area of each triangle w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the area of each triangle with measures given. $$b=6, c=4 \sqrt{3}, \alpha=30^{\circ}$$

Short Answer

Expert verified
The area of the triangle is \(6\sqrt{3}\).

Step by step solution

01

Understand the Given Values

We are given the base of the triangle, \(b = 6\), side \(c = 4\sqrt{3}\), and \(\alpha = 30^\circ\), which is the angle between sides \(b\) and \(c\). Our goal is to find the area of the triangle using this information.
02

Recall the Area Formula

Recall the formula to find the area of a triangle when two sides and the included angle are known: \(A = \frac{1}{2}bc\sin\alpha\). Here, \(b\) and \(c\) are the sides of the triangle, and \(\alpha\) is the included angle.
03

Substitute the Values

Substitute the given values into the formula: \(b = 6\), \(c = 4\sqrt{3}\), and \(\alpha = 30^\circ\). This gives us: \[A = \frac{1}{2} \times 6 \times 4\sqrt{3} \times \sin(30^\circ)\] Since \(\sin(30^\circ) = \frac{1}{2}\), the expression becomes:\[A = \frac{1}{2} \times 6 \times 4\sqrt{3} \times \frac{1}{2}\].
04

Simplify the Expression

Simplify the multiplicative expression:\[A = \frac{1}{2} \times 6 \times 4\sqrt{3} \times \frac{1}{2} = 3 \times 2\sqrt{3} = 6\sqrt{3}\].Thus, the area of the triangle is \(6\sqrt{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry
Trigonometry is an essential branch of mathematics that explores the relationships between the angles and sides of triangles. It is especially useful when it comes to solving problems involving triangles, such as finding unknown angles or calculating areas. In trigonometry, we use specific functions — sine, cosine, and tangent — to express and relate these elements easily.
When we deal with non-right triangles, like in our given exercise, these trigonometric functions are pivotal. They allow us to calculate important properties through formulas and rules. For example, using the sine function, we can find the area of a triangle when we know two sides and the included angle, as seen in the given solution.
For those studying trigonometry, it's important to understand the unit circle, which helps define sine and other functions for all angle measures, ensuring these concepts are not only applicable to right triangles but any type of triangle.
Sine Rule
The sine rule is a vital concept within trigonometry, used to relate the angles and sides of a triangle. It states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all three sides. In formulaic terms, it is expressed as:
  • \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
Here, \(a\), \(b\), and \(c\) are the sides of the triangle, while \(A\), \(B\), and \(C\) are the respective opposite angles.
This rule is particularly useful when you need to find missing sides or angles in non-right angled triangles. In our given problem, however, the sine rule allows us to work with the sine function, which is crucial for calculating the area of the triangle.
Understanding and applying the sine rule can significantly simplify many trigonometric problems and proves to be an indispensable tool in both academic and real-world applications of geometry.
Included Angle
An included angle in a triangle is the angle formed between two sides of the triangle. It can be crucial in determining different properties of the triangle, such as its area or its shape.
In our specific problem, the included angle \(\alpha = 30^\circ\) allows us to use the formula for area:
  • \(A = \frac{1}{2}bc\sin\alpha\)
This formula is especially helpful when two sides of the triangle and the angle between them (the included angle) are known. It provides a straightforward way to calculate the triangle's area without needing other side lengths or angles.
By recognizing the included angle, you can simplify complex problems and find solutions efficiently. This approach is not only practical for solving homework exercises but also in many fields where geometry is applied, such as engineering and architecture.

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Most popular questions from this chapter

Find the measure (in degrees, minutes, and nearest seconds) of a central angle \(\theta\) that intercepts an are on a circle with indicated radius \(r\) and are length \(s .\) With the TI calculator commands \([\text { ANGLE }]\) and \([\text { DMS }],\) change to degrees, minutes, and seconds. $$r=78.6 \mathrm{cm}, s=94.4 \mathrm{cm}$$

Determine whether each statement is true or false. If you are given the measures of one side and one acute angle of a right triangle, you can solve the right triangle.

Explain the mistake that is made. Solve the triangle \(a=6, b=2,\) and \(c=5\). Solution: Step 1: Find \(\beta\) Apply the Law of Cosines. \(b^{2}=a^{2}+c^{2}-2 a c \cos \beta\) Solve for \(\beta\) \(\beta=\cos ^{-1}\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)\) Let \(a=6, b=2\) \(c=5 . \quad \beta \approx 18^{\circ}\) Step 2: Find \(\alpha\) \(\begin{array}{ll}\text { Apply the Law } & \frac{\sin \alpha}{a}=\frac{\sin \beta}{b} \\ \text { of sines. } & a\end{array}\) Solve for \(\alpha\) \(\alpha=\sin ^{-1}\left(\frac{a \sin \beta}{b}\right)\) Let \(a=6, b=2\) and \(\beta=18^{\circ}\) \(\alpha \approx 68^{\circ}\) Step 3: Find \(\gamma\) \(\alpha+\beta+\gamma=180^{\circ}\) $$ \begin{aligned} 68^{\circ}+18^{\circ}+\gamma &=180^{\circ} \\ \gamma & \approx 94^{\circ} \end{aligned} $$ \(a=6, b=2, c=5, \alpha \approx 68^{\circ}, \beta \approx 18^{\circ},\) and \(\gamma \approx 94^{\circ}\) This is incorrect. The longest side is not opposite the largest angle. What mistake was made?

Let \(A, B,\) and \(C\) be the lengths of the three sides with \(X, Y,\) and \(Z\) as the corresponding angle measures in a triangle. Write a program using a TI calculator to solve each triangle with the given measures. $$B=45, C=57, \text { and } X=43^{\circ}$$

In calculus, the value of \(F(b)-F(a)\) of a function \(F(x)\) at \(x=a\) and \(x=b\) plays an important role in the calculation of definite integrals. Find the exact value of \(F(b)-F(a)\). $$F(x)=\frac{\cot x-4 \sin x}{\cos x}, a=\frac{\pi}{4}, b=\frac{\pi}{3}$$

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