10.04 Solving for sides in right triangles
Recall our 3 trigonometric ratios that relate an angle and two sides of a right triangle together.
$\sin\theta$sinθ = $\frac{Opposite}{Hypotenuse}$OppositeHypotenuse
$\cos\theta$cosθ = $\frac{Adjacent}{Hypotenuse}$AdjacentHypotenuse
$\tan\theta$tanθ = $\frac{Opposite}{Adjacent}$OppositeAdjacent
If we know $2$2 side lengths or an angle and a side length, then we can then find any other part of that triangle using these trigonometric ratios.
Finding side lengths
If we know $2$2 sides and want to find the third, we would use Pythagorean theorem.
For a right triangle $\triangle ABC$△ABC with hypotenuse $c$c, the following is true about its side lengths:
$a^2+b^2=c^2$a2+b2=c2
If we know $1$1 side length and $1$1 acute angle, we would use one of the trigonometric ratios.
The most common mistake is when the wrong ratio is used. We have to remember the ratios and the sides that apply to those ratios. For most students the mnemonic SOHCAHTOA can be a great help.
Worked examples
Question 1
In the given triangle $\theta=25^\circ$θ=25° and the hypotenuse measures $12.6$12.6. Solve for the length $b$b to two decimal places.
Think: We need to identify the sides we have and want with respect to the angle given. Here we can see that we have the hypotenuse (H) and we want $b$b, which is opposite (O) the angle. This means we have OH - so the trig ratio we need to use here is sine.
Do:
| $\sin\theta$sinθ | $=$= | $\frac{O}{H}$OH |
State the formula for the correct ratio |
| $\sin25^\circ$sin25° | $=$= | $\frac{b}{12.6}$b12.6 |
Fill in the given values |
| $b$b | $=$= | $12.6\times\sin25^\circ$12.6×sin25° |
Multiply both sides by 12.6 |
| $b$b | $=$= | $5.32$5.32 |
Simplify |
Question 2
Find the length of the hypotenuse ($c$c) in the diagram, where the angle $36^\circ$36° and the side length of $4.8$4.8 are given, to two decimal places.
Think: We need to identify the sides we have and want with respect to the angle given. Here we can see that we want the hypotenuse (H) and we have a side length of $4.8$4.8, which is adjacent (A) the angle. This means we have AH - so the trig ratio we need to use here is cosine.
Do:
| $\cos\theta$cosθ | $=$= | $\frac{A}{H}$AH |
State the formula for the correct ratio |
| $\cos36^\circ$cos36° | $=$= | $\frac{4.8}{c}$4.8c |
Fill in the given values |
| $c\times\cos36^\circ$c×cos36° | $=$= | $4.8$4.8 |
Multiply both sides by $c$c, (this step is often skipped) |
| $c$c | $=$= | $\frac{4.8}{\cos36^\circ}$4.8cos36° |
Divide both sides by $\cos36^\circ$cos36° |
| $c$c | $=$= | $5.93$5.93 |
Simplify |
Question 3
Find the length of the unknown side, when the angle is $66^\circ$66° and the indicated side length is $7.3$7.3, to two decimal places.
Think: We need to identify the sides we have and want with respect to the angle given. Here we can see that we want the adjacent side ($A$A) and we have a side length of $7.3$7.3, which is opposite ($O$O) the angle. This means we have $OA$OA - so the trig ratio we need to use here is tangent.
Do:
| $\tan\theta$tanθ | $=$= | $\frac{O}{A}$OA |
State the formula for the correct ratio |
| $\tan66^\circ$tan66° | $=$= | $\frac{7.3}{a}$7.3a |
Fill in the given values |
| $a$a | $=$= | $\frac{7.3}{\tan66^\circ}$7.3tan66° |
Cross multiply (really multiplying by $a$a and dividing by $\tan66^\circ$tan66°) |
| $a$a | $=$= | $3.25$3.25 |
Simplify |
The vast majority of the time, we will be asked to round to a specified number of decimal places. However, if the acute angle is $30^\circ$30°, $45^\circ$45° or $60^\circ$60°, we can use the exact values. Our answer may involve a radical.
| $30^\circ$30° | $45^\circ$45° | $60^\circ$60° | |
|---|---|---|---|
 |
 |
|
|
| sin | $\sin30^\circ=\frac{1}{2}$sin30°=12 | $\sin45^\circ=\frac{\sqrt{2}}{2}$sin45°=√22 | $\sin60^\circ=\frac{\sqrt{3}}{2}$sin60°=√32 |
| cos | $\cos30^\circ=\frac{\sqrt{3}}{2}$cos30°=√32 | $\cos45^\circ=\frac{\sqrt{2}}{2}$cos45°=√22 | $\cos60^\circ=\frac{1}{2}$cos60°=12 |
| tan | $\tan30^\circ=\frac{\sqrt{3}}{3}$tan30°=√33 | $\tan45^\circ=1$tan45°=1 | $\tan60^\circ=\sqrt{3}$tan60°=√3 |
Practice questions
Question 4
Find the value of $f$f, correct to two decimal places.
Question 5
Find the value of $x$x, the side length of the parallelogram, to the nearest centimeter.
A parallelogram with its angle at upper-right corner labeled as 52 degrees, indicating its measure. Other angles of the parallelogram are not labeled. Its top side is labeled 32 cm, indicating its length. Its left side is labeled $x$x cm, indicating its unknown length. A perpendicular internal segment is drawn from the upper-left corner to the bottom side of the parallelogram, thus creating a right triangle on the left side. The bottom side of the parallelogram, which is parallel to the top side as indicated by the double arrowheads, is cut into two parts: the left part with no measurement indicated and the right part which measures 11 cm, as labeled.
In the triangle, the sides are the internal segment, the left side of the parallelogram, and the left part of the bottom side of the parallelogram. A square symbol is shown at the corner where the internal segment and the bottom side intersect to indicate that it is a right angle. The left side of the parallelogram labeled $x$x cm acts as the hypotenuse of the triangle. The internal segment is the vertical leg of the triangle. The left part of the bottom side of the parallelogram acts as horizontal leg of the triangle.
Question 6
Find the length of side $h$h.
