Solve the following triangles
i.e. find all unknown sides and angles
$\u25b5$ ABC in which
$\widehat{\mathrm{A}}={70}^{\circ}$ ;
$b=4$ and
$c=9$
$\u25b5$ XYZ in which
$\widehat{\mathrm{Y}}={112}^{\circ}$ ;
$x=2$ and
$y=3$
$\u25b5$ RST in which RS
$=2$ ; ST
$=3$ and RT
$=5$
$\u25b5$ KLM in which KL
$=5$ ; LM
$=10$ and KM
$=7$
$\u25b5$ JHK in which
$\widehat{\mathrm{H}}={130}^{\circ}$ ; JH
$=13$ and HK
$=8$
$\u25b5$ DEF in which
$d=4$ ;
$e=5$ and
$f=7$
Find the length of the third side of the
$\u25b5$ XYZ where:
$\widehat{\mathrm{X}}=71,{4}^{\circ}$ ;
$y=3,42$ km and
$z=4,03$ km
;
$x=103,2$ cm;
$\widehat{\mathrm{Y}}=20,{8}^{\circ}$ and
$z=44,59$ cm
Determine the largest angle in:
$\u25b5$ JHK in which JH
$=6$ ; HK
$=4$ and JK
$=3$
$\u25b5$ 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
$\u25b5DEF$ , the area is given by:
$A=\frac{1}{2}DE\xb7EFsin\widehat{E}=\frac{1}{2}EF\xb7FDsin\widehat{F}=\frac{1}{2}FD\xb7DEsin\widehat{D}$
In order show that this is true for all triangles, consider
$\u25b5ABC$ .
The area of any triangle is half the product of the base and the perpendicular height. For
$\u25b5ABC$ , this is:
$A=\frac{1}{2}c\xb7h.$ However,
$h$ can be written in terms of
$\widehat{A}$ as:
$h=bsin\widehat{A}$ So, the area of
$\u25b5ABC$ is:
$A=\frac{1}{2}c\xb7bsin\widehat{A}.$
Using an identical method, the area rule can be shown for the other two angles.
Find the area of
$\u25b5$ ABC:
$\u25b5$ ABC is isosceles, therefore AB
$=$ AC
$=7$ and
$\widehat{\mathrm{C}}=\widehat{\mathrm{B}}={50}^{\circ}$ . Hence
$\widehat{\mathrm{A}}={180}^{\circ}-{50}^{\circ}-{50}^{\circ}={80}^{\circ}$ . Now we can use the area rule to find the area:
Q is a ship at a point 10 km due South of another ship P. R is a lighthouse on the coast such that
$\widehat{\mathrm{P}}=\widehat{\mathrm{Q}}={50}^{\circ}$ .
Determine:
the distance QR
the shortest distance from the lighthouse to the line joining the two ships (PQ).
WXYZ is a trapezium (WX
$\parallel $ XZ) with WX
$=3$ m; YZ
$=1,5$ m;
$\widehat{\mathrm{Z}}={120}^{\circ}$ and
$\widehat{\mathrm{W}}={30}^{\circ}$
Determine the distances XZ and XY.
Find the angle
$\widehat{\mathrm{C}}$ .
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:
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.
The correction, to one hundredth of a degree, to the plane's heading (or direction).
ABCD is a trapezium (ie. AB
$\parallel $ CD). AB
$=x$ ;
$\mathrm{B}\widehat{\mathrm{A}}\mathrm{D}=\mathrm{a}$ ;
$\mathrm{B}\widehat{\mathrm{C}}\mathrm{D}=\mathrm{b}$ and
$\mathrm{B}\widehat{\mathrm{D}}\mathrm{C}=\mathrm{c}$ .
Find an expression for the length of CD in terms of
$x$ ,
$a$ ,
$b$ and
$c$ .
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
$\mathrm{X}\widehat{\mathrm{Y}}\mathrm{Z}$ .
If XY
$=x$ and
$\mathrm{X}\widehat{\mathrm{Y}}\mathrm{Z}=\theta $ , show that XZ
$=x\sqrt{2(1-cos\theta )}$ .
Calculate XZ (to the nearest kilometre) if
$x=240$ km and
$\theta ={132}^{\circ}$ .
Find the area of WXYZ (to two decimal places):
Find the area of the shaded triangle in terms of
$x$ ,
$\alpha $ ,
$\beta $ ,
$\theta $ and
$\phi $ :
Questions & Answers
where we get a research paper on Nano chemistry....?
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
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How this robot is carried to required site of body cell.?
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analytical skills graphene is prepared to kill any type viruses .
Anam
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