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Stage 1 - Stage 3

Applications to Geometry

Lesson

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.  

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.

Triangle BCD is drawn with a small green square at vetex D indicating that it is a right triangle. Side BC, the hypotenuse, measures 20 cm. Side CD, the base, is labeled as y cm. And side BD, is the height, is labeled as x cm. From vertex B, a perpendicular line segment is drawn with a length of 3 cm and has its end labeled as A. Then, a diagonal line is drawn connecting A to vertex D and has a measurement of 13 cm. Triangle ADB is formed cojoining triangle BDC.

  1. What is the value of $x$x? Give your answer correct to two decimal places.

  2. 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.

 

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