# 11.1 Graphs, trigonometric identities, and solving trigonometric  (Page 12/12)

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The other cases can be proved in an identical manner.

Find $\stackrel{^}{\mathrm{A}}$ :

1. $\begin{array}{ccc}\hfill {a}^{2}& =& {b}^{2}+{c}^{2}-2bdcos\stackrel{^}{\mathrm{A}}\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}cos\stackrel{^}{\mathrm{A}}& =& \frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\hfill \\ & =& \frac{{8}^{2}+{5}^{2}-{7}^{2}}{2·8·5}\hfill \\ & =& 0,5\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}\stackrel{^}{\mathrm{A}}& =& arccos0,5={60}^{\circ }\hfill \end{array}$

## The cosine rule

1. Solve the following triangles i.e.  find all unknown sides and angles
1. $▵$ ABC in which $\stackrel{^}{\mathrm{A}}={70}^{\circ }$ ; $b=4$ and $c=9$
2. $▵$ XYZ in which $\stackrel{^}{\mathrm{Y}}={112}^{\circ }$ ; $x=2$ and $y=3$
3. $▵$ RST in which RS $=2$ ; ST $=3$ and RT $=5$
4. $▵$ KLM in which KL $=5$ ; LM $=10$ and KM $=7$
5. $▵$ JHK in which $\stackrel{^}{\mathrm{H}}={130}^{\circ }$ ; JH $=13$ and HK $=8$
6. $▵$ DEF in which $d=4$ ; $e=5$ and $f=7$
2. Find the length of the third side of the $▵$ XYZ where:
1. $\stackrel{^}{\mathrm{X}}=71,{4}^{\circ }$ ; $y=3,42$  km and $z=4,03$  km
2. ; $x=103,2$  cm; $\stackrel{^}{\mathrm{Y}}=20,{8}^{\circ }$ and $z=44,59$  cm
3. Determine the largest angle in:
1. $▵$ JHK in which JH $=6$ ; HK $=4$ and JK $=3$
2. $▵$ PQR where $p=50$ ; $q=70$ and $r=60$

## The area rule

The Area Rule

The area rule applies to any triangle and states that the area of a triangle is given by half the product of any two sides with the sine of the angle between them.

That means that in the $▵DEF$ , the area is given by: $A=\frac{1}{2}DE·EFsin\stackrel{^}{E}=\frac{1}{2}EF·FDsin\stackrel{^}{F}=\frac{1}{2}FD·DEsin\stackrel{^}{D}$

In order show that this is true for all triangles, consider $▵ABC$ .

The area of any triangle is half the product of the base and the perpendicular height. For $▵ABC$ , this is: $A=\frac{1}{2}c·h.$ However, $h$ can be written in terms of $\stackrel{^}{A}$ as: $h=bsin\stackrel{^}{A}$ So, the area of $▵ABC$ is: $A=\frac{1}{2}c·bsin\stackrel{^}{A}.$

Using an identical method, the area rule can be shown for the other two angles.

Find the area of $▵$ ABC:

1. $▵$ ABC is isosceles, therefore AB $=$ AC $=7$ and $\stackrel{^}{\mathrm{C}}=\stackrel{^}{\mathrm{B}}={50}^{\circ }$ . Hence $\stackrel{^}{\mathrm{A}}={180}^{\circ }-{50}^{\circ }-{50}^{\circ }={80}^{\circ }$ . Now we can use the area rule to find the area:

$\begin{array}{ccc}\hfill A& =& \frac{1}{2}cbsin\stackrel{^}{\mathrm{A}}\hfill \\ & =& \frac{1}{2}·7·7·sin{80}^{\circ }\hfill \\ & =& 24,13\hfill \end{array}$

## The area rule

Draw sketches of the figures you use in this exercise.

1. Find the area of $▵$ PQR in which:
1. $\stackrel{^}{\mathrm{P}}={40}^{\circ }$ ; $q=9$ and $r=25$
2. $\stackrel{^}{\mathrm{Q}}={30}^{\circ }$ ; $r=10$ and $p=7$
3. $\stackrel{^}{\mathrm{R}}={110}^{\circ }$ ; $p=8$ and $q=9$
2. Find the area of:
1. $▵$ XYZ with XY $=6$  cm; XZ $=7$  cm and $\stackrel{^}{\mathrm{Z}}={28}^{\circ }$
2. $▵$ PQR with PR $=52$  cm; PQ $=29$  cm and $\stackrel{^}{\mathrm{P}}=58,{9}^{\circ }$
3. $▵$ EFG with FG $=2,5$  cm; EG $=7,9$  cm and $\stackrel{^}{\mathrm{G}}={125}^{\circ }$
3. Determine the area of a parallelogram in which two adjacent sides are 10 cm and 13 cm and the angle between them is ${55}^{\circ }$ .
4. If the area of $▵$ ABC is 5000 m ${}^{2}$ with $a=150$  m and $b=70$  m, what are the two possible sizes of $\stackrel{^}{\mathrm{C}}$ ?

## Summary of the trigonometric rules and identities

 Pythagorean Identity Ratio Identity ${cos}^{2}\theta +{sin}^{2}\theta =1$ $tan\theta =\frac{sin\theta }{cos\theta }$
 Odd/Even Identities Periodicity Identities Cofunction Identities $sin\left(-\theta \right)=-sin\theta$ $sin\left(\theta ±{360}^{\circ }\right)=sin\theta$ $sin\left({90}^{\circ }-\theta \right)=cos\theta$ $cos\left(-\theta \right)=cos\theta$ $cos\left(\theta ±{360}^{\circ }\right)=cos\theta$ $cos\left({90}^{\circ }-\theta \right)=sin\theta$ Sine Rule Area Rule Cosine Rule $\mathrm{Area}=\frac{1}{2}\mathrm{bc}cos\mathrm{A}$ ${a}^{2}={b}^{2}+{c}^{2}-2bccosA$ $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$ $\mathrm{Area}=\frac{1}{2}\mathrm{ac}cos\mathrm{B}$ ${b}^{2}={a}^{2}+{c}^{2}-2accosB$ $\mathrm{Area}=\frac{1}{2}\mathrm{ab}cos\mathrm{C}$ ${c}^{2}={a}^{2}+{b}^{2}-2abcosC$

## Exercises

1. Q is a ship at a point 10 km due South of another ship P. R is a lighthouse on the coast such that $\stackrel{^}{\mathrm{P}}=\stackrel{^}{\mathrm{Q}}={50}^{\circ }$ . Determine:
1. the distance QR
2. the shortest distance from the lighthouse to the line joining the two ships (PQ).
2. WXYZ is a trapezium (WX $\parallel$ XZ) with WX $=3$  m; YZ $=1,5$  m; $\stackrel{^}{\mathrm{Z}}={120}^{\circ }$ and $\stackrel{^}{\mathrm{W}}={30}^{\circ }$
1. Determine the distances XZ and XY.
2. Find the angle $\stackrel{^}{\mathrm{C}}$ .
3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has been flying ${3}^{\circ }$ off course. At this point the plane is 500 km from Johannesburg. The direct distance between Cape Town and Johannesburg airports is 1 552 km. Determine, to the nearest km:
1. The distance the plane has to travel to get to Cape Town and hence the extra distance that the plane has had to travel due to the pilot's error.
2. The correction, to one hundredth of a degree, to the plane's heading (or direction).
4. ABCD is a trapezium (ie. AB $\parallel$ CD). AB $=x$ ; $\mathrm{B}\stackrel{^}{\mathrm{A}}\mathrm{D}=\mathrm{a}$ ; $\mathrm{B}\stackrel{^}{\mathrm{C}}\mathrm{D}=\mathrm{b}$ and $\mathrm{B}\stackrel{^}{\mathrm{D}}\mathrm{C}=\mathrm{c}$ . Find an expression for the length of CD in terms of $x$ , $a$ , $b$ and $c$ .
5. A surveyor is trying to determine the distance between points X and Z. However the distance cannot be determined directly as a ridge lies between the two points. From a point Y which is equidistant from X and Z, he measures the angle $\mathrm{X}\stackrel{^}{\mathrm{Y}}\mathrm{Z}$ .
1. If XY $=x$ and $\mathrm{X}\stackrel{^}{\mathrm{Y}}\mathrm{Z}=\theta$ , show that XZ $=x\sqrt{2\left(1-cos\theta \right)}$ .
2. Calculate XZ (to the nearest kilometre) if $x=240$  km and $\theta ={132}^{\circ }$ .
6. Find the area of WXYZ (to two decimal places):
7. Find the area of the shaded triangle in terms of $x$ , $\alpha$ , $\beta$ , $\theta$ and $\phi$ :

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what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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da
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Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
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please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
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Professor
I think
Professor
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Rafiq
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Damian
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LITNING
scanning tunneling microscope
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Rafiq
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Mahi
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Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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write examples of Nano molecule?
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
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Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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