Solving Triangles

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

• 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

$\text{Area} = \frac12 bc \sin(A)= \frac12 ac \sin(B) = \frac12 ab \sin(C)$

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

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By the cosine rule, we have $c^2 = a^2 + b^2 - 2ab \cos(C)$. Substituting all the relevant information, we have $c^2 = \frac{391}7$. Since $c$ represents a side length of a triangle, $c> 0$ 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 two of its side lengths being 7 and 8, what is the sine of the angle in between these two side lengths?

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Using the area formula, $\text{Area} = \frac12 bc \sin(A)\implies 20 = \frac12 \cdot 7 \cdot 8 \sin(A).$ Solving for the sine of the angle gives $\frac57.\ _\square$

The area of an acute triangle is 10, and two side lengths of the triangle are 6 and 7. What is the cosine of the angle in between these two sides?

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Let the angle between the two sides be $A.$ Then the area of the triangle is $10 =\frac12\times 6 \times 7 \times \sin(A),$ which gives $\sin(A) = \frac{10}{21}.$

By applying the Pythagorean identity, $\sin^2(A) + \cos^2(A) = 1,$ so $\cos(A) = \sqrt{1- \left(\frac{10}{21}\right)^2} = \frac{\sqrt{341}}{21}.$ $_\square$

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

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?

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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 $\frac{1234}{(abc)^2} = \sin(A) \sin(B) \sin(C).\ _\square$