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Solve the triangle. In other words, find the measurements of all unknown sides and angles. If two triangles are possible, solve for both. $$B C=42, A C=37 \text { and } A B=26$$.

Short Answer

Expert verified
The triangle has angles A ≈ 37.45°, B ≈ 64.33°, C ≈ 78.22°. Only one triangle is possible.

Step by step solution

01

Identify Triangle Type

We have a triangle with sides given: \(BC = 42\), \(AC = 37\), and \(AB = 26\). Since all three sides are given, we will use the cosine rule to find an angle, confirming if the triangle is right, acute, or obtuse.
02

Calculate Angle Using Cosine Rule

We apply the cosine rule to find angle \(A\), which states: \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \). Here, \(a = 26\), \(b = 37\), \(c = 42\). Substituting into the cosine rule: \[ \cos A = \frac{37^2 + 42^2 - 26^2}{2 \times 37 \times 42} \]\[ \cos A = \frac{1369 + 1764 - 676}{3108} = \frac{2457}{3108} \approx 0.79 \]Calculating \(A\) gives \(A \approx \arccos(0.79) \approx 37.45^\circ \).
03

Use Sine Rule to Find Another Angle

With angle \(A\) known, use the sine rule to find \(B\) or \(C\). Using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} \), compute angle \(B\):\[ \sin B = \frac{37 \times \sin(37.45^\circ)}{26} \approx 0.897 \]\[ B \approx \arcsin(0.897) \approx 64.33^\circ \].
04

Calculate Remaining Angle

Using the angle sum property of a triangle (\(A + B + C = 180^\circ\)), solve for \(C\): \[ C = 180^\circ - 37.45^\circ - 64.33^\circ = 78.22^\circ \]
05

Conclusion Regarding Possibility of Second Triangle

Check for the possibility of a second triangle by considering if another configuration could produce valid angles with given sides. The largest angle \(C\) calculated is opposite the largest side; hence no alternate solutions or ambiguous cases exist.

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Key Concepts

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

Cosine Rule
The cosine rule is a fundamental concept for solving triangles, particularly useful when you know all three sides but none of the angles. It allows us to find one of the angles when the other two sides and opposite side length are known. This is incredibly helpful in non-right-angled triangles.In our exercise, the cosine rule for finding angle \(A\) is given as:\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]Here, \(a\) is the side opposite angle \(A\), and \(b\) and \(c\) are the other two sides. By plugging the lengths of the sides into the formula, we can solve for \(\cos A\) and consequently use the inverse cosine function to find the angle \(A\). A critical thing to note is the relationship this rule reveals between the angles and sides, helping to deduce whether a triangle is acute, right-angled, or obtuse based on the calculated angle.
Sine Rule
The sine rule is another important tool in triangle problem-solving, allowing us to find unknown angles or sides in a triangle when we already have some of them.The formula states:\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]In the given exercise, once we've determined angle \(A\) using the cosine rule, the sine rule helps us find another angle \(B\). Knowing \(A\) and its corresponding side \(a\), we can solve for \(B\) using:\[ \sin B = \frac{b \times \sin A}{a} \]This method is particularly useful because it simplifies finding additional unknown angles when one angle and its opposite side are known.
Triangle Types
Understanding the types of triangles is vital when solving triangle problems, as each type offers different properties and rules. Triangles can be classified based on:
  • Angles: Right, acute, or obtuse.
  • Sides: Equilateral, isosceles, or scalene.
In our exercise, by using the cosine rule, we initially check the triangle type based on its angles. A right triangle has one 90-degree angle, whereas an acute triangle has all angles less than 90 degrees. An obtuse triangle has one angle greater than 90 degrees. Knowing a triangle's type can determine which rules and formulas are applicable, simplifying the problem-solving process further.
Angle Sum Property
The angle sum property is a simple yet powerful concept that states the sum of all interior angles in any triangle is always \(180^\circ\).In solving triangles, after calculating two angles using the cosine and sine rules, we can easily find the third angle using this property. This makes it a convenient way to complete our angle calculations.For the given problem, once we find angles \(A\) and \(B\), we compute angle \(C\) by subtracting the sum of \(A\) and \(B\) from \(180^\circ\):\[ C = 180^\circ - A - B \]This step is crucial as it ensures the triangle is correctly solved, with all angles neatly adding up to the expected total. Understanding this property not only helps compute missing angles but also verifies the accuracy of your calculations.

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Most popular questions from this chapter

For each question a) sketch a right triangle corresponding to the given trigonometric function of the acute angle \(\theta,\) b) find the exact value of the other five trigonometric functions, and c) use your GDC to find the degree measure of \(\theta\) and the other acute angle (approximate to 3 significant figures). $$\sin \theta=\frac{\sqrt{7}}{4}$$

For triangle \(A B C\) we can express its area, \(\mathbf{A},\) as \(\mathbf{A}=\frac{1}{2} a b \sin C .\) The cosine rule can be used to write the expression \(c^{2}=a^{2}+b^{2}-2 a b \cos C\). a) Using these two expressions show that \(16 A^{2}=4 a^{2} b^{2}-\left(a^{2}+b^{2}-c^{2}\right)\) \(\cdot\) Hint: use the Pythagorean identity \(\sin ^{2} C+\cos ^{2} C=1\) b) The perimeter of the triangle is equal to \(a+b+c .\) Let \(s\) be the semiperimeter, that is \(s=\frac{a+b+c}{2}\). Using the result from a) and that $$2 s=a+b+c, \text { show that } 16 \mathbf{A}^{2}=2 s(2 s-2 c)(2 s-2 a)(2 s-2 b)$$ c) Finally, show that the result in b) gives \(A=\sqrt{s(s-a)(s-b)(s-c)} .\) This notable result expresses the area of a triangle in terms of only the length of its three sides. Although quite possibly known before his time, the formula is attributed to the ancient Greek mathematician and engineer, Heron of Alexandria (ca. \(10-70 \mathrm{AD}\) ) and is thus called Heron's formula. The first written reference to the formula is Heron's proof of it in his book Metrica, written in approximately \(60 \mathrm{AD}\)

For each question a) sketch a right triangle corresponding to the given trigonometric function of the acute angle \(\theta,\) b) find the exact value of the other five trigonometric functions, and c) use your GDC to find the degree measure of \(\theta\) and the other acute angle (approximate to 3 significant figures). $$\tan \theta=2$$

In triangle \(F G H, F G=12 \mathrm{cm}, F H=15 \mathrm{cm},\) and \(\angle G\) is twice the size of \(\angle H .\) Find the approximate degree measure of \(\angle H\) to three significant figures.

A 1.62 -metre tall woman standing 3 metres from a streetlight casts a 2 -metre long shadow. What is the height of the streetlight?

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