Solve for sides or angles I
So far we have found unknown side lengths using Pythagoras' theorem and then looked at 3 special ratios that we can use to find unknown sides or angles in right-angled triangles.
Pythagoras' theorem: $a^2+b^2=c^2$a2+b2=c2, where $c$c is the hypotenuse
$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse = $\frac{O}{H}$OH
$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse = $\frac{A}{H}$AH
$\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent =$\frac{O}{A}$OA
Problem solving in trigonometry can be in finding unknowns like we have already been doing, using trigonometry in real world applications or in solving geometrical problems like these.
Examples
Question 1
Find $x$x in the following geometrical diagram,
Think: In order to find $x$x, I will need to identify some other measurements along the way. My problem solving strategy will be
1. Find length $AC$AC using trig ratio sine
2. Find length $ED$ED, $\frac{AC}{3}$AC3
3. Find length $x$x, using trig ratio sine
Do:
1. Find length $AC$AC using trig ratio sine
| $\sin23^\circ$sin23° | $=$= | $\frac{43.6}{AC}$43.6AC |
| $AC$AC | $=$= | $\frac{43.6}{\sin23^\circ}$43.6sin23° |
| $AC$AC | $=$= | $111.59$111.59 |
2. Find length $ED$ED, $\frac{AC}{3}$AC3
$ED=\frac{111.59}{3}$ED=111.593
$ED=37.2$ED=37.2
3. Find length $x$x, using trig ratio sine
| $\sin35.6^\circ$sin35.6° | $=$= | $\frac{x}{37.2}$x37.2 |
| $x$x | $=$= | $37.2\times\sin35.6^\circ$37.2×sin35.6° |
| $x$x | $=$= | $21.65$21.65 |
Question 2
Consider the following diagram.
What is the value of $x$x? Give your answer correct to two decimal places.
Using the value of $x$x you got from part (a), find the value of $y$y correct to two decimal places.
Question 3
Find the length of the unknown side, $x$x, in the given trapezium.
Give your answer correct to two decimal places.
A right trapezoid $ABDC$ABDC is depicted as suggested by the two adjacent right angles $\angle BAC$∠BAC or $\angle CAB$∠CAB on vertex $9$9 and $\angle DCA$∠DCA or $\angle ACD$∠ACD on vertex $7$7. Side $AB$AB or $BA$BA and Side $DC$DC or $CD$CD are the parallel sides of the trapezoid and Side $AB$AB or $BA$BA is longer than side $DC$DC or $CD$CD. Side $CA$CA or $AC$AC measures $9$9 units and is perpendicular to the two parallel sides. Side $CA$CA or $AC$AC is the base of the figure. Side $AB$AB or $BA$BA is measured as $13$13 units. Side $DC$DC or $CD$CD is measured as $7$7 units. Side $BD$BD or $DB$DB is labeled as $x$x units.