3.02 Finding unknown sides

When we consider the angles and sides of a triangle together, we instinctively know there is a relationship between them.

For example, we know that the largest angle in a triangle is opposite the longest side, and the smallest angle is opposite the shortest side.

But we can be even more specific about the relationship between the angles and sides of right-angled triangles.

We start by relating each side of a right-angled triangle to a given angle:

As the angle $x$x changes, the side lengths change as well. But the ratios of sides remain constant.

For example, here are three right-angled triangles where the other two angles are $45^\circ$45° each. This means they are also isosceles triangles, with two equal sides.

As the triangles get larger, the side lengths increase, but the ratio of $\frac{\text{Opposite }}{\text{Adjacent }}$Opposite Adjacent ​ from the $45^\circ$45° angle always stays the same. In each triangle the ratio is $\frac{1}{1}$11​, $\frac{2}{2}$22​ and $\frac{3}{3}$33​ and all of these ratios simplify to $1$1.

There are other ratios that remain constant as well, and we can explore these constant ratios for any angle, not just $45^\circ$45°. We call these trigonometric ratios, and the basic ratios have names $\sin$sin, $\cos$cos and $\tan$tan:

Every angle has a fixed $\sin$sin, $\cos$cos and $\tan$tan ratio, so they are programmed into our calculators.

For example, we can use our calculators to find the following values:

• $\sin30^\circ=0.5$sin30°=0.5   (The side opposite $30^\circ$30° is half the length of the hypotenuse)
• $\cos25^\circ=0.9$cos25°=0.9 (The side adjacent to $25^\circ$25° is $0.9$0.9 times the length of the hypotenuse)
• $\tan82^\circ=7.1$tan82°=7.1  (The side opposite $82^\circ$82° is about $7$7 times the length of the adjacent side)

Having these fixed trig ratios allows us to solve for other unknowns.

Finding side lengths

If we know one angle and one side length in a right-angled triangle, then we can then find any other side in the same triangle. 

The most important part of the process is to correctly identify the ratio that relates the angle and the sides we are interested in. These examples shows us the process:

When the unknown is in the numerator

In this right-angled triangle we have:

• a $25^\circ$25° angle
• the hypotenuse equal to $12.6$12.6
• the side opposite $25^\circ$25° unknown

 

The two sides are the opposite side and the hypotenuse, so we can use the $\sin$sin ratio.

$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse ​		The $\sin$sin ratio.
$\sin25^\circ=\frac{b}{12.6}$sin25°=b12.6​		Substituting values into the ratio.
$b=12.6\times\sin25^\circ$b=12.6×sin25°		Rearranging the equation to solve for $b$b.
$b=5.32$b=5.32  ($2$2 d.p.)		Using our calculator to solve for $b$b.

The first step is the most important. We need to identify which ratio will help us relate the angle and sides. Once we have this, we just need to solve the equation correctly.

When the unknown is in the denominator

 

In this right-angled triangle we have:

• a $36^\circ$36° angle
• the hypotenuse unknown
• the adjacent side length equal to $4.8$4.8

 

The two sides are the adjacent side and the hypotenuse, so we can use the $\cos$cos ratio.

$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse ​		The $\cos$cos ratio.
$\cos36^\circ=\frac{4.8}{c}$cos36°=4.8c​		Substituting values into the ratio
$c\times\cos36^\circ=4.8$c×cos36°=4.8		Rearranging the equation to solve for $c$c.
$c=\frac{4.8}{\cos36^\circ}$c=4.8cos36°​		Rearranging again to make $c$c the subject.
$c=5.93$c=5.93  (2 d.p.)		Using our calculator to solve for $c$c.

 

 

Examples

Question 1

Find the value of $a$a in the given triangle to two decimal places.

First we need to identify the the unknown side and the given sides, with respect to the angle in the triangle. Here, we want the adjacent side (A) and we are given the side opposite the angle (O), so we use the $\tan$tan ratio.

$\tan\theta$tanθ	$=$=	$\frac{Opposite}{Adjacent}$OppositeAdjacent​
$\tan66^\circ$tan66°	$=$=	$\frac{7.3}{a}$7.3a​
$a$a	$=$=	$\frac{7.3}{\tan66^\circ}$7.3tan66°​
$a$a	$=$=	$3.25$3.25 mm ($2$2 d.p.)

 

Question 2

Question 3

Question 4