Question

For questions $1-10$ solve $\mathrm{△}\mathrm{ABC}$ given $B={42}^{\circ },C={93}^{\circ },\mathrm{BC}=13\mathrm{cm}$

Short answer

Answer: In the triangle, angle A is ${45}^{\circ }$, side AB (opposite angle B) is approximately 8.77 cm, and side AC (opposite angle A) is approximately 9.15 cm.

Step-by-step solution

Find Angle A

To find angle A, we can use the fact that the sum of the angles in a triangle is always equal to ${180}^{\circ }$. So: $A={180}^{\circ }-B-C$ Substituting the given values, we get: $A={180}^{\circ }-{42}^{\circ }-{93}^{\circ }={180}^{\circ }-{135}^{\circ }={45}^{\circ }$

Find Side AC (a)

Now, we will use the Law of Sines to find the length of side AC (a): $\frac{a}{\sin (A)}=\frac{c}{\sin (C)}$ We have c=13, A=45°, and C=93°. We need to solve for a: $a=\frac{c\cdot \sin (A)}{\sin (C)}$ Substituting the values, we get: $a=\frac{13\cdot \sin ({45}^{\circ })}{\sin ({93}^{\circ })}$ Now, we can use a calculator to find the values of sine: $a\approx \frac{13\cdot 0.7071}{0.9952}\approx 9.15\mathrm{cm}$

Find Side AB (b)

Similarly, we will use the Law of Sines to find the length of side AB (b): $\frac{b}{\sin (B)}=\frac{c}{\sin (C)}$ We have c=13, B=42°, and C=93°. We need to solve for b: $b=\frac{c\cdot \sin (B)}{\sin (C)}$ Substituting the values, we get: $b=\frac{13\cdot \sin ({42}^{\circ })}{\sin ({93}^{\circ })}$ Now, we can use a calculator to find the values of sine: $b\approx \frac{13\cdot 0.6691}{0.9952}\approx 8.77\mathrm{cm}$ So, the triangle $\mathrm{△}ABC$ has side lengths a=9.15 cm, b=8.77 cm, and c=13 cm, with angles A=${45}^{\circ }$, B=${42}^{\circ }$, and C=${93}^{\circ }$.

Key concepts

Law of Sines

The Law of Sines is a critical formula used for solving various problems involving triangles, especially when dealing with non-right triangles. It provides a relationship between the lengths of sides and the sines of their opposite angles. The formula is expressed as:

$$\frac{a}{\sin (A)}=\frac{b}{\sin (B)}=\frac{c}{\sin (C)}$$
Where $a$, $b$, and $c$ are the lengths of the sides of the triangle, and $A$, $B$, and $C$ are the angles opposite these sides, respectively. This formula is especially useful in situations where we know either two angles and one side (AAS or ASA cases), or two sides and an opposite angle (SSA case). In the given exercise, the Law of Sines is adeptly applied to find the unknown sides of the triangle after determining the third angle using the triangle angle sum theorem.

Triangle Angle Sum Theorem

Understanding the triangle angle sum theorem is fundamental when we wish to solve the angles of a triangle effectively. This theorem states that the sum of all three angles in a triangle is always equal to 180 degrees. The formula can be written as:

$$A+B+C={180}^{\circ }$$
In the context of the original exercise, the first step was to use this theorem to find the unknown angle $A$, since angles $B$ and $C$ were given. Knowing that the sum of angles must be 180 degrees, the problem simplifies to a basic subtraction. Thus, determining angle $A$ is essential to employ the Law of Sines and resolve the remaining unknowns of the triangle. This theorem is pivotal to numerous geometric calculations and serves as a cornerstone for solving triangles.

Sine Function

The sine function is one of the primary trigonometric functions and is key to solving many problems involving right and non-right triangles. It's defined for an angle as the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right triangle.

The general sine function can be written as:

$$\sin (\theta )=\frac{\text{opposite}}{\text{hypotenuse}}$$
However, in the context of the Law of Sines, which applies to any triangle, the sine function relates a side and its opposite angle. Calculating the sine of angles, as done in the exercise, allows us to work out unknown side lengths or angles when coupled with the Law of Sines. The exercise uses the sine values for angles $A$ and $B$, and compares them with the sine of angle $C$ to find missing sides $a$ and $b$. The sine function's versatility makes it an invaluable tool in various branches of mathematics and physics.