A triangle can be of 2 types:
- right angle
- non right angle
With the right angle triangle, one angle is 90 degrees. The sides of the triangles are labelled or named with respect to an angle (not the 90) in the triangle.
- Identify the angle
- Longest side hyp
- opp is opposite to the angle
- adj is adjacent to the angle
Anything to be found can be evaluated using sin, cos, tan or pythagoras theorem.
With a non right angle triangle
- label the vertices
- label the angles
- Label the length associated with the angle
Use sine rule or cosine rule
In trig questions it is important to identify all the triangles in the figure and draw the separately
- DEFG is a trapezium with DX a perpedicular line to GF. DE = 10cm, DG = 13cm, Angle EFX and DXF are right angles. Find the length DX. The area of the trapezium.
- ABC and PCD are right angle triangles. Angle ABC = 40 degrees, AB = 10 cm, PD = 8 cm and BD = 15cm. Angles BCA and DCP are right angles. Find the length BC the angle PDC.
- A plane takes off at an angle of elevation 17 degrees to the ground. After 25 seconds the plane has travelled a horizontal distance of 1400. Calculate the height of the plane above the ground after 25 seconds.
- STW is a triangle with the angle STW is 52 degrees. ST = 5cm, TW is 9cm. Calculate the lenght = SW and the area of triangle STW.
I dont know how to solve the first question as i feel there is inadequate info given...for example an angle for the triangle DGX? please correct me if im wrong or if im missing something...but i dont know how to approach it...
ReplyDeletealso another length could be given for the same triangle...
ReplyDeletequestion three:
ReplyDeletewe have the angle 17 degrees and since the vertical height is wanted, we can say that the triangle is a right angled triangle. we can also now say that we know two angles,17 degrees and 90 degrees (the right angle) and can find the last angle :- 180 - (90 + 17) = 73 degrees.
let the vertical height = x;
using the sine rule:
a/sinA = b/sinB = c/sinC;
x/sin 17 = 1400/sin73
x = (1400/sin73) x (sin17)
x = 428.02 km
therefore vertical height after 25 secs = 428.02 km
question four:
ReplyDeleteis this a right angled triangle or not?
assuming it is not:
using cosine rule, a^2 = b^2 + c^2 - 2bc cosA;
SW^2 = ST^2 + WT^2 - 2(ST)(WT) cos angle STW
SW^2 = 5^2 + 9^2 - 2(5)(9)cos52
= 25 + 81 - 90 cos 52
= 106 - 90 cos 52
SW^2 = 50.59
SW = 7.11 cm
question four cont'd...
ReplyDeleteWT is the length of the total base.
in order to find the area of the triangle we need to find the height of the triangle.
splitting the base in half, we now have a base of 4.5 cm and a hypothenus of 5 cm.
now we can find the vertical height.
let the height be equal to 'h';
using Pythagoras' Theorem, a^2 = b^2 + c^2;
5^2 = 4.5^2 + h^2
25 = 20.25 + h^2
h^2 = 25 - 20.25
h^2 = 4.75
h = 2.18 cm
question four cont'd...
ReplyDeletenow we can find the area of the triangle.
area of triangle = (base x height)/2
= (9 x 2.18)/2
area of triangle = 9.81 cm^2
hmmmm....isn't angles of elevation deals with right angle triangles....correct me if i am wrong bornagain 16..thanx
ReplyDeletebornagain 16 u use the sine rule in number 3 and it deals with angle of elevation so if u can explain why you use the sine rule that will be nice. :)
ReplyDeleteThis comment has been removed by the author.
ReplyDelete4)since the question did not mention the triangle beeing a right angle triangle.we assume the triangle to a non right angle triangle and use the cosine rule since we have two sides and a given angle.
ReplyDelete4) calculations
ReplyDeleteusing the cosine rule to caluclate the length of SW.rule is a^2=b^2+c^2-2ba cosine 52
SW= 5^2+9^2-2(5)(9)cosine 52
SW= 25+81-90*0.615
SW=106-55.35
SW=50.65
SW=7.12 cm
4)to calculate the area of the triangle use the formula 1/2 AB sin c
ReplyDeletearea=1/2 AB sin c
area= 1/2 (5)(7.12) sin 52
area=1/2 35.6 *0.778
area= 17.8 * 0.778
area=14.03cm
bornagain16 first you assume that number 4 is not a right angle triangle but then to calculate the height you use phythagoras theorm..can you check it over please.thank you and tell me if i am goin correct because i getting a different answer to you. :)
ReplyDeleteyes thats true...i dont know how i ended up using the sine rule for that question. i even stated that it is a right angled triangle!!! thanks for showing me that XOXO S...
ReplyDeletethere is adequate information. remember a trapezium is made up of triangles and a square..so if u have a length of a square u can find the other lengths.all the lengths in a square are equal.
ReplyDeletesure bornagain16 . always a pleasure..lolz :)
ReplyDelete1) length of DX is 10cm..DE is 10cm so therefore DX is 10 cm well.
ReplyDeletefor question 3 the information given is that the angle of elevation= 17, the horizontal distance of the plane in 25 seconds=1400, the vertical distance of the plane in 25 seconds= ?.
ReplyDeletesince it is a right angled triangle the angles in the triangle add up to 180. therefore 180-(90+17)=73.
we can use tan rule to solve this equation.
tanᶿ= adj/opp
tan17= h/1400
1400tan17= h
h=428.02
another method we can use is the sin rule:
ReplyDeletea/sinA= b/sinB= c/sinC
1400/sin73= h/sin17
cross multiply the equation to get:
1400 sin17= h sin73
1400 sin17/ sin73= h
h= 428.02
boom!!!! and there you have it..
let the height be equal to 'h';
ReplyDeleteusing Pythagoras' Theorem, a^2 = b^2 + c^2;
5^2 = 4.5^2 + h^2
25 = 20.25 + h^2
h^2 = 25 - 20.25
h^2 = 4.75
h = 2.18 cm
yes you can also use sin as commented by hellomoto
ReplyDelete(4) seeing that we have the sides ST= 5cm and TW= 9cm and the angle between them STW=52o, we can use cosine rule to find the area of the non- right angle triangle.
ReplyDeletecosine rule: c^2= a^2 + b^2 - 2(a)(b)cosC
c^2= 5^2 + 9^2 - 2(5)(9)Cos52
c^2= 25 + 81 -90 cos52
c= √25 + 81 -90 cos52
c=7.11
therefore the length of SW= 7.11cm
Because triangle STW is not a right angled triangle its area can be found using the formula: 1/2absinC
ReplyDelete=1/2absinC
=1/2 (5)(9)sin52
=1/2 (45sin52)
=1/2 (35.460)
=17.73cm^2
4)
ReplyDeleteusing cosine rule:
known values:
st=5cm
tw=9cm
angle stw=52 degrees
cosine rule: a^2=b^2+c^2-2bc cos A
SW^2 = 5^2 + 9^2 - 2(5)(9)cos52
= 25 + 81 - 90 cos 52
= 106 - 90 cos 52
SW^2 = 50.59
SW = 7.11 cm
4)
ReplyDeletequestion did not state whether triangle was right angled or not...so i just assumed that it was not.in order to use the cos rule!!
4)The question did not say if the triangle was a right angle triangle.I assumed the triangle to be a non right angle triangle and so I use the cosine rule since there is two sides and a given angle.
ReplyDeletetw=9cm
ReplyDeletest=5cm
angle stw=52 degrees
using the cosine rule a^2 = b^2 + c^2 - 2bc cosA
SW^2 = 5^2 + 9^2 - 2(5)(9)cos52
= 25 + 81 - 90 cos 52
= 106 - 90 cos 52
SW^2 = 50.59
SW = 7.11 cm
I agree with bornagain 16
ReplyDeleteat least one more lenght or an angle is needed to solve question one
wow i tried sooooooo much with question 1 but damn its impossible to figure out...u really do need to get angle DGX or another length to figure this one out
ReplyDeletehey XOXO S don't know if you know but well obviously you dont know...not all trapeziums are amde up of triangles with SQUARES...it can also be made up of triangles with RECTANGLES and somtimes even RHOMBUSES...cool huh???? lol
ReplyDeleteI've read through all the workings you guys did for #3 and they all have the same sine rule method! and someone else also used the tan rule to solve the problem...why????? this is a very very very simple problem, with an even simplier working....
ReplyDeleteThis is how I would solve this problem here goes:
ReplyDeleteThe question says that the plane took off at an angle of elevation of 17degrees so all this means is that you have to draw the horizontal and then construct an angle of 17degrees to this horizontal. Then it states that after 25 secs. the plane travelled 1400 this would just be the hypotenuse in this problem, given this they want you to find the height of the plane above ground(horizontal), the height would be the opposite in this problem...
From all of this the problem can be solved like this:
sin@ = opp/hyp
sin17= opp/1400
opp = 1400sin17
opp = 409.32
TRIG TEASER QUESTION
ReplyDeleteFor a right angle triangle if sinx = 3/4
find lenght of all the sides and also the area. IT IS UNBELEIVABLY EASY GOOD LUCK!
Since two sides and an angle is included the you can use the cosine rule to solve #4
ReplyDeletea^2 = b^2 + c^2 - 2(b)(c)cosA
a^2 = (5^2)+(9^2) - 2(5)(9)cos52
a^2 = (25 + 81) - (90cos52)
a^2 = 106 - 55.41
a^2 = 50.59
a = square root of 50.59
a = 7.11
since the triangle is not right angle, the area can be found using the formula 1/2 absin@
ReplyDeleteie. 1/2(7.11)(5)sin52
= 17.775sin52
= 14.01
REAPER grrrrrr.......
ReplyDeletelol@ REAPER ok omg...i figured it out...since sinx = 3/4 it means that the hypotenuse is 4 and the opposite side is 3 since sinx = opp/hyp
ReplyDeleteright??? lol and then using these sides the adjacent side can be found using Pythagoras' theorem ie. a^2 = b^2 + c^2 and the adjacent would work out to be 5
a^2 = 4^2 + 3^2
a^2 = 16 + 9
a^2 = 25
a = root 25
a = 5
now given all three sides and knowing that it is a right angle triangle the area can ba found using the formula 1/2bh
ie. 1/2(5)(3)
1/2(15)
7.5
I have a question for you guys to try...its kinda challenging so if you think your good enough try it!!!!
ReplyDeleteQUESTIIN:
From point A, the angle of elevation to the top a tall building is 20 degrees. On walking 80 m towards the building the angle of elevation is now 23 degrees. How tall is the building?
GOOD LUCK GUYS!!!!!
hey guys here's a kinda challenging question for the trigs loving bloggers.
ReplyDeleteQUESTION:
From point A, the angle of elevation to the top a tall building is 20 degrees. On walking 80 m towards the building the angle of elevation is now 23 degrees. How tall is the building?
HAVE FUN AND GOOD LUCK!!!!!!111
ok i did the question already...but i need a way to draw diagrams here!!!i cant explain my method properly without diagrams...
ReplyDelete4. Seeing that you have two sides and the included angle, the cosine rule can be used to find the length.
ReplyDeletecosine rule: c^2= a^2 + b^2 - 2(a)(b)cosC
c^2= 5^2 + 9^2 - 2(5)(9)Cos52
c^2= 25 + 81 -90 cos52
c= √25 + 81 -90 cos52
c=7.1
Length 7.1
OMG!!! SERIOUSLY DIAGRAMS WOULD HELP SOOOOOOOOOO MUCH!!!!!! *SIGH* it would make explaining our methods sooo much easier.
ReplyDeletetrigs
ReplyDeleteFind the values of
a. sin(-30)
b. sec(-150)
c. cot(-300)
using the sin rule:
ReplyDeletea/sinA= b/sinB= c/sinC
1400/sin73= h/sin17
cross multiply the equation to get:
1400 sin17= h sin73
1400 sin17/ sin73= h
h= 428.02
question 1 makes my brain scream, there should be more info.
ReplyDeletecan someone help me out with No. 2
ReplyDeletefor no. 3 i didnt use the sine rule cuz i think angles of elevation deals with right angles. i worked it noramal and got the same ans as born again.
ReplyDeletemy workin is
ReplyDeletetan@= opp/adj
opp = tan@* adj
= tan17 * 1400
= 428.02
therefore the height is 428.02
please correct me if im incorrect
quetion 4.
ReplyDeletetw=9cm
st=5cm
angle stw=52 degrees
using the cosine rule a^2 = b^2 + c^2 - 2bc cosA
SW^2 = 5^2 + 9^2 - 2(5)(9)cos52
= 25 + 81 - 90 cos 52
= 106 - 90 cos 52
SW^2 = 50.59
SW = 7.11 cm
correct me if i'm wrong!!!!
the use of diagrams would really make a difference!!!
ReplyDeletethe angle 17 degrees and since the vertical height is wanted, we can say that the triangle is a right angled triangle. we can also now say that we know two angles,17 degrees and 90 degrees (the right angle) and can find the last angle :- 180 - (90 + 17) = 73 degrees.
ReplyDeletelet the vertical height = x;
using the sine rule:
a/sinA = b/sinB = c/sinC;
x/sin 17 = 1400/sin73
x = (1400/sin73) x (sin17)
x = 428.02 km
therefore vertical height after 25 secs = 428.02 km
4)
ReplyDeletearea=1/2 AB sin c
area= 1/2 (5)(7.12) sin 52
area=1/2 35.6 *0.778
area= 17.8 * 0.778
area=14.03cm
tw=9cm
ReplyDeletest=5cm
angle stw=52 degrees
using the cosine rule a^2 = b^2 + c^2 - 2bc cosA
SW^2 = 5^2 + 9^2 - 2(5)(9)cos52
= 25 + 81 - 90 cos 52
= 106 - 90 cos 52
SW^2 = 50.59
SW = 7.11 cm
that number 4
ReplyDeletearea of triangle stw is.......
ReplyDelete=1/2absinC
=1/2 (5)(9)sin52
=1/2 (45sin52)
=1/2 (35.460)
=17.73cm^2
no. 3 let x be the height of the plane
ReplyDeleteusing the sine rule:
a/sinA = b/sinB = c/sinC;
x/sin 17 = 1400/sin73
x = (1400/sin73) x (sin17)
x = 428.02 km
therefore vertical height after 25 secs = 428.02 km
Question 3
ReplyDeletewe have the angle 17 degrees and since the vertical height is wanted, we can say that the triangle is a right angled triangle. we can also now say that we know two angles,17 degrees and 90 degrees (the right angle) and can find the last angle :- 180 - (90 + 17) = 73 degrees.
let the vertical height = x;
using the sine rule:
a/sinA = b/sinB = c/sinC;
x/sin 17 = 1400/sin73
x = (1400/sin73) x (sin17)
x = 428.02 km
vertical height after 25 secs = 428.02 km
let the height be equal to 'h';
ReplyDeleteusing Pythagoras' Theorem, a^2 = b^2 + c^2;
5^2 = 4.5^2 + h^2
25 = 20.25 + h^2
h^2 = 25 - 20.25
h^2 = 4.75
h = 2.18 cm
we have the angle 17 degrees and since the vertical height is wanted, we can say that the triangle is a right angled triangle. we can also now say that we know two angles,17 degrees and 90 degrees (the right angle) and can find the last angle :- 180 - (90 + 17) = 73 degrees.
ReplyDeletelet the vertical height = x;
using the sine rule:
a/sinA = b/sinB = c/sinC;
x/sin 17 = 1400/sin73
x = (1400/sin73) x (sin17)
x = 428.02 km
vertical height after 25 secs = 428.02 km
4.a^2 = b^2 + c^2 - 2(b)(c)cosA
ReplyDeletea^2 = (5^2)+(9^2) - 2(5)(9)cos52
a^2 = (25 + 81) - (90cos52)
a^2 = 106 - 55.41
a^2 = 50.59
a = square root of 50.59
a = 7.11
Because triangle STW is not a right angled triangle its area can be found using the formula: 1/2absinC
ReplyDelete=1/2absinC
=1/2 (5)(9)sin52
=1/2 (45sin52)
=1/2 (35.460)
=17.73cm^2
i always seem to get confused between the adjacent side and the opposite sides on a triangle.
ReplyDeleteThe opposite side is the side that is directly opposite the angle you are taking into consideration.
ReplyDeletethen there's the hypotenus which is the longest side
and the remaining sise is the adjacent side
4). STW= 52 degrees
ReplyDeleteST= 5 cm
TW= 9 cm
using the cosine rule you could find SW
SW^2=TW^2+ST^2-2(ST)(TW)COS T
SW^2=9^2+5^2-2(5)(9)COS 52
SW^2=106-55.41
SW^2=50.59
SW=SQRT50.59
SW=7.11
to find the area of the triangle you have to use HERON'S FORMULA
i.e. A=sqrts(s-a)(s-b)(s-c)
and s=semiperimeter
s=a+b+c/2
s=5+9+7.11/2=10.56
A=sqrt10.56(10.56-5)(10.56-9)(10.56-7.11)
A=sqrt10.56(5.56)(1.56)(3.45)
A=sqrt315.997
A=17.78 cm^2
3). using tan you could find the height the plane was travelling.
ReplyDeletetan angle= o/a
tan17=o/1400
o=tan17*1400
o=428.02
the height the plane was travelling at is 428.02.
2). BCA AND PCD IS RIGHT ANGLES
ReplyDeleteBA=10 cm
DP=8 cm
BD=15 cm
to find BC you have to use the sin angle=o/h
sin50=o/10
o=sin50*10
o= 7.60cm
so therefore BC is 7.66cm
NEED SOME HELP TO FIND THE ANGLE PDC.
(MISS,ANYONE)
PLEASE HELP…………………IS THIS POSSIBLE???
ReplyDeleteFrom a point 435 feet from the base of a building it is observed that the angle of elevation to the top of the building is 24° and the angle of elevation to the top of a flag pole on top the building is 27°.
PLEASE HELP…………………IS THIS POSSIBLE???
ReplyDeleteI am 10 feet away from a man lift that holds the worker repairing the electricity poll. My angle of elevation is 50 degrees my eyes are 5 feet from the ground and the man lift is 30 feet. How tall is the worker?
HOW TO WORK THIS???
ReplyDeletey = 3x , between x = 0 , and x= 3, for
a) n =3 (∆x = 1) , (b) n = 10 (∆x = 0.3)
Reaper's trig teaser question:
ReplyDeletesinx=3/4
sinx=opp/hyp
therefore opp=3 (side b) and hyp=4 (side c)
pythagoras' theorem states a^2+b^2=c^2
a^2+3^2=4^2
a^2=4^2-3^2
=16-9=7
a=root of 7=2.65
area of tri. =1/2 (b*h)
=1/2 (2.65*3)=3.98
#2
ReplyDeletecos=adj/hyp
cos40=adj/10
.766=adj/10
adj=.766(10)
BC=7.6
#2
ReplyDeletesin40=opp/hyp
.643=opp/10
opp=.643(10)
AC=6.4
#2
ReplyDeletesin tata=opp/hyp
sin tata=6.4/8
tata=(0.8)-sin
angle D=53degre
#3
ReplyDeletetan17=opp/adj
opp=(1400)tan17
opp=(1400)86.6
opp=428.02
#4
ReplyDeleteSW=the squar root of 9^2-5^2
SW=the squar root of 81-25
SW=7.48
#4
ReplyDeleteA=B*H
A=5*7.48
A=37.4
me no figure out question one!!!!! lolz
ReplyDeleteassuming our diagram is a right angled triangle
ReplyDelete180-(90 + 17) = 73 degrees.(for our 3rd angle)
let the vertical height = x;
using the sine rule:
a/sinA = b/sinB = c/sinC;
x/sin 17 = 1400/sin73
x = (1400/sin73) x (sin17)
x = 428.02 km
vertical height after 25 secs = 428.02 km
no3.
ReplyDeletetan = opp/adj
tan17= x/1400
1400tan17=x
x= 428
height of plane after 25 sec is 428