# Solving Triangles

To solve problems on this page, you should be familiar with:

- Sine Rule

\[ \frac a{\sin(A) } = \frac b{\sin(B)} = \frac c{\sin(C) } \]- Cosine Rule

\[ a^2=b^2+c^2-2bc\cos(A) \\ b^2 = a^2+c^2-2ac\cos(B) \\ c^2= a^2+b^2-2ab\cos(C) \]- Area Of Triangle
\[\text{Area} = \frac12 bc \sin(A)= \frac12 ac \sin(B) = \frac12 ab \sin(C) \]

## Problem Solving - Basic

When we want to *solve* a triangle, this means that we want to find all/some of the unknown lengths and angles of the triangle.

A triangle \(ABC\) has side length \(a = 4, b= 9\) and \(C = \cos^{-1} \left(\frac47\right) \). What is the length of the third side, \(c\)?

Solution: By Cosine Rule, we have \(c^2 = a^2 + b^2 - 2ab \cos(C) \). Substituting all the relevant information, we have \(c^2 = \frac{391}7 \). And because \(c\) represent a side length of a triangle, then \(c> 0 \) only, so \(c = \sqrt{ \frac{391}7} \approx 7.474\). \(\ \square\).

If the area of a triangle is given to be 20 unit squared with 2 of its side length as 7 and 8 respectively. What is the sine of the angle in between these two given side lengths?

Solution: Area is given to be \(\text{Area} = \frac12 bc \sin(A) \), given \(20 = \frac12 \cdot 7 \cdot 8 \sin(A) \), solving for the sine of the angle gives \( \frac57 \ \square \).

## Problem Solving - Intermediate

If the area of an acute angle triangle is given to be 10 with two sides given to be 6 and 7 respectively. What is the cosine of the angle in between these two sides?

Solution: Let angle between the two sides to be denoted as \(A\). Then we have the area given to be \(10 =\frac12 6 \times 7 \times \sin(A) \). Solving for \(A\) yields \(\sin(A) = \frac{10}{21} \).By applying the Pythagorean Identity, \( \sin^2(A) + \cos^2(A) = 1\), then \(\cos(A) = \sqrt{1- \left(\frac{10}{21}\right)^2} = \frac{\sqrt{341}}{21} \). \( \ \square\).

Note that we are only taking the positive root only because cosine of an acute angle is strictly positive.

If the numerical value of the area of a triangle is given to be \( \frac{\sqrt[3]{1234}}2 \). What is the relationship between the product of the sides of the triangle to the product of sines of the angles of the triangle?

Solution: Let \(D\) denote the area of the triangle, then \(\frac{\sqrt[3]{1234}}2 = D = \frac12 ab \sin(C) = \frac12 ac \sin(B) = \frac12 bc \sin(A) \). Then,\[ D^3 = \frac{1234}8 = \frac18 (abc)^2 \sin(A) \sin(B) \sin(C) \]

So the relationship in question is simply \( \frac{1234}{(abc)^2} = \sin(A) \sin(B) \sin(C) \)

A semi-circle is inscribed within an equilateral triangle such that the diameter of the semi-circle is centred on one side of the triangle and the arc is tangent to the other two sides. If each side of the triangle is of length 4 cm, then what is the diameter of the semi-circle (in cm)?

Give your answer to 3 decimal places.

## Problem Solving - Advanced

In a \(\Delta ABC \), if

\[a^4 + b^4 + c^4 = 2c^2(a^2 +b^2) \]

then the possible measure(s) of \(\angle C\) could be?

**Details and Assumptions**

\(a,b,c\) are the sides of \(\Delta ABC\) opposite to \(\angle A , \angle B, \angle C\) respectively.

\(\angle C\) is measured in

**radians**.

**Cite as:**Solving Triangles.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/solving-triangles/