2D & 3D Geometry

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Non-right-angled triangles

The formulas we used in the previous section applied only to right triangles. You might be wondering how useful these formulas and rules even are if they only apply to such a limited selection of problems, and you'd be right to do so! Luckily, we can actually generalize the trig ratios and other formulas for area and side length to the case of non-right triangles.

Let's start by considering a non-right triangle which is split into right triangles, and seeing how our old formulas apply.

Discussion

The diagram shows a non-right triangle with vertices A, B, and C. A line perpendicular to side AB drops down from the draggable vertex C, splitting ΔABC into two right triangles. The angle ∠BAC and length of side AC will update dynamically as you move C.

Try to come up with a formula for the length of the dashed red line. Can you use this length to write a general formula for the area of ΔABC?

Let ∠BAC=θ and AC=b. The dashed line CD is the altitude from C onto AB. In right‐triangle ADC we have

CD=ACsinθ=bsinθ

Since the area of △ABC is one half the base AB times the height CD,

Area=21AB⋅CD=21AB(bsinθ)=21b(AB)sinθ

If we write AB=c and remember θ=∠BAC, this becomes

Area=21bcsin(∠BAC)

In other words, for any triangle the area equals one half the product of two sides times the sine of the angle between them:

A=21(side1)(side2)sin(included angle)

We can use this idea to find the area of any triangle:

Area of non-right-angled triangles

When we know two side lengths and the angle between the two sides, we can find the area even if we don't directly know the height by using the fact that

sinC=ah⇒h=asinC

Thus

A=21(bh)=21absinC📖

Example

Triangle ABC has [AB]=5, [BC]=3, and ∠C=35°. Find the area of the triangle.

A=215⋅3⋅sin(35°)≈4.30

Exercise

Find the area of the triangle below.

Select the correct option

We can similarly extend the trig ratios to non-right triangles, even without splitting the triangle into two right ones.

Discussion

The diagram shows a non-right triangle with vertices A, B, and C. A line perpendicular to side AB drops down from the vertex C, splitting ΔABC into two right triangles.

The formula we found above for area applies to any pair of sides and the angle between them, i.e.

A=21absinC=21bcsinA=21acsinB

Can you rearrange this expression to find a relationship between the side lengths a, b, and c?

Starting from the area formula for any triangle,

A=21absinC=21bcsinA=21acsinB,

equate the first two expressions:

21absinC=21bcsinA⟹absinC=bcsinA⟹asinC=csinA⟹asinA=csinC.

Next equate the first and third expressions:

21absinC=21acsinB⟹absinC=acsinB⟹bsinC=csinB⟹bsinB=csinC.

Hence all three ratios are equal:

asinA=bsinB=csinC.

Equivalently, inverting each gives

sinAa=sinBb=sinCc.

This relationship is a crucial formula known as the sine rule.

Sine rule

The previously found formula for area

A=21absinC📖

applies to any pair of sides and the angle between them. Since the area is the same no matter which sides we use:

21absinC=21bcsinA=21acsinB

Multiplying everything by 2 and dividing by abc:

csinC=asinA=BsinB🚫

Flipping the numerator and denominator gives the form that appears in the formula booklet:

sinAa=sinBb=sinCc📖

The sine rule is primarily used when we know two angles and a side. When we know two sides and an angle, the version with angles in the numerator is easier to work with.

Example

Find the side x in the triangle below.

The sine rule works with angles and the side opposite them, so we need the angle opposite x. Since the angles in a triangle add to 180°:

C=180°−70°−41°=69°

The sine rule tells us that

sin(70°)6=sin(69°)x⇒x=sin(70°)6sin(69°)≈5.96

Exercise

Find the length x.

Select the correct option

The sine rule does work with every triangle -- but due to the symmetric nature of angles and the sine function, there are some cases where the answer may be ambiguous.

Ambiguous Case of the Sine rule

The ambiguous case of the sine rule is a consequence of the symmetry of the sin function:

sin(180°−A)=sin(A)

Therefore, if a given triangle has a specific sin(A) and it is not specified whether A is acute / obtuse, then there are two possible lengths for the side opposite A:

When we type sin−1(x) into the calculator, it will always return an acute angle. To find the corresponding obtuse angle, we take

180°−sin−1(x)

Example

In triangle ABC, the angle A=30° and the sides [BC]=10 and [AC]=12 are given. Find

the angle B,
the length [AB].

The angle B is opposite [AC], and A is opposite [BC], so:

sin(30°)10=sinB12⇒sinB=0.6

This is where the symmetry of sin comes into play: B=sin−1(0.6)=36.9° or B=180°−36.9°=143°

The resulting length [AB] can be found by finding the angle C=180°−30°−(36.9° or 143°)=113° or 6.87°.

Thus sinC=0.920 or 0.120. Then

sinC[AB]=sinA[BC]⇒[AB]=20sinC=18.4 or 2.39

Exercise

In triangle ABC, ∣AB∣=4, ∣BC∣=3 and ∠BAC=40∘. Find ∠BCA.

Select the correct option

Discussion

We can derive another generalization of a trig ratio, this time cosine, by generalizing the Pythagorean theorem. Let's walk through it step by step.

First, split up △ABC into two right triangles, △ACD and △BCD, each with height h. Can you write expressions for a and b in terms of these smaller right triangles?
The equations from part (a) have a common variable. Through combining and rearranging them, show that
AD2−DB2=(AD−DB)(AD+DB).
Now use this equation, along with the fact that AD−DB=AD−(c−AD), to find an equation for AD.
Use your equation from part (c) to write an expression for cosθ in terms of AD.
Finally, solve the equation cosθ=2bcb2+c2−a2 for a2.

Part (a)

Drop the altitude CD to the base AB at D. This splits △ABC into two right triangles, △ACD and △BCD, each with height h.

By the Pythagorean theorem in △ACD and △BCD,

b2=AD2+h2,a2=DB2+h2.

Part (b)

Subtract a2=DB2+h2 from b2=AD2+h2 to eliminate h2:

b2−a2=AD2−DB2.

Notice the right‐hand side is a difference of squares, so

AD2−DB2=(AD−DB)(AD+DB).

Part (c)

Since AD+DB=AB=c, we have

b2−a2=(AD−DB)c.

But AD−DB=AD−(c−AD)=2AD−c. Therefore

b2−a2=(2AD−c)c⟹2AD−c=cb2−a2⟹2AD=cb2−a2+c2⟹AD=2cb2−a2+c2.

Part (d)

In △ACD, the angle at A is θ. The cosine of θ is the adjacent side over the hypotenuse, so

cosθ=bAD=2bcb2−a2+c2=2bcb2+c2−a2.

Part (e)

Rearranging the equation,

a2=b2+c2−2bccosθ

which is known as the Law of Cosines. Geometrically, it “corrects” the Pythagorean theorem by subtracting the product 2bccosθ that accounts for the angle θ being non–right.

Cosine rule

The cosine rule is a generalization of Pythagoras' theorem for non-right-angled triangles. It states that

c2=a2+b2−2abcosC📖

The cosine rule is primarily used when we

know two sides and the angle between them, and want to find the third side,
know all three sides and want to find an angle.

Example

Triangle ABC has [AB]=3, [BC]=4 and [AC]=5. Find the angle C.

The angle C is between the sides [BC] and [AC] so

[AB]2 9 cosC=[BC]2+[AC]2−2[BC]⋅[AC]cosC =16+25−2⋅4⋅5cosC =4032⇒C≈36.9°

Exercise

Find the length x.

Select the correct option

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