In Trigonometry, you have learnt about SINE Rule. In using SINE rule to solve some problems, there is a situation in which the solution becomes AMBIGUOUS.
Task:
1. When does a problem becomes AMBIGUOUS when you apply SINE rule?
2. Craft a question with its solution demonstrating AMBIGUOUS case when you apply SINE rule.
Total marks: 10
Due date: Friday, 3rd August 2007
PS: No copying of example given in textbook.
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33 comments:
i dun reali understand the question =) can u explain more clearer?
wad is it about huh ?
-alicia
Q1:When does a problem becomes AMBIGUOUS when you apply SINE rule?
A: The problem becomes ambiguous when the triangle is a congruent as it is SSA and therefore, it may not be determined from the given conditions.
------------------------------------------
Q2:Craft a question with its solution demonstrating AMBIGUOUS case when you apply SINE rule.
A: In Triangle ABC, a=12cm, c=18cm & Angle A=30 deg. Find Angle C.
1st Solution:
12/sin30 = 18/sinC
sinC= 0.75
C= sin-1(0.75)
=48.6
2nd Solution:
180deg-48.6deg
=131.4 deg
Therefore, Angle C = 48.6deg or 131.4deg.
Note: deg means degrees
.: This question and solution demonstrates Ambiguous case when applying the SINE rule.
Hi people, Miss Yong has taught to all of you about this ambigous case last Wednesday in class. Pls check with her if you are not clear about this. Deadline is Monday, 6th August 2007.
Mr Malek
QN1: If two sides, a and b, and a non included angle* in a ▲ named ▲ABC
~ b sin* < a < b, there are two possible triangles
~ when a > b, there exists only 1 ▲
QN2: In ▲PQR, q= 7cm, r= 5cm and
< Q = 47 degrees. Find < P, < R and the side p.
Solution
7/Sin47 = 5/Sin < R
Sin< R = 0.5224
< R = 31.5 degrees or
148.5 degrees (rejected)
Gregory
forgot to say although i do the work i still dun really understand.. pls dun explain too fast.. and more clearer pls..
Gregory
i'm not confident enough to answer haha
qns 1: when the triangle can be drawn in two different shapes and ways (maybe correct or wrong not sure)
qns2:too blur to think(omg!!! faint...)
Q1..(ans);
questions becomes ambiguous when e triangle is congruent as it is SSA and that is the reason that it may not be determined from the given information.
Q2..(ans);
Triangle XYZ, x=20 z=24 and angle X=40 degree.
ans..
20/sin40 = 24/sinZ
sinZ=0.771( 3 dp )
Z= sin-1(0.771)
Z=50.4 ( 3 dp )
junyan ..
Question 1) The problem becomes ambiguous when the triangle is a congruent as it is SSA and therefore, it may not be determined from the given conditions.
If two sides, a and b, and a non included angle* in a triangle named triangle ABC
~ b sin* < a < b, there are two possible triangles
~ when a > b, there exists only 1 triangle
Question 2 ) in triangle XYZ , x = 13cm , z = 20cm and angle X = 35degrees . find the angle of Z.
Solution :
13/sin35 = 20/sinZ
13sinZ = 20sin35
13sinZ = 11.47
sinZ = 11.47/13
sinZ = 0.8824
Z = 61.9degrees or 180-61.9
= 118.1degree
-Alicia , copyrighted
Q1 : ... Ambiguous case happens when the fact SSA ( two sides and a non included angle) is not a condition for congruent triangles.
Q2: Given ▲ ABC with a=10cm , c=25cm , < A = 20deg. Find < B , < C
Answear:
a/sinA = c/sinC
10/sin = 25/sin C
=> sin C = 0,86
=> C= 59,3 or C = 120,7 .
when < C = 59,3 => < B = 100,7
when < C = 120,7 => < B = 39,3
Vic
Q1 ANSWER:
When we know one angle of the triangle,and its opposite side is shorter than the line beside it,then it will become AMBIGUOUS when we apply SINE rule.
Q2 ANSWER:
in the triangleABC,B=31',c=18 and b=10,find the angleC.
SOLUTION ONE----
sin 31'/10=sin C/18
sinC=0.927
sin-1=0.927
angle C=68.0 DEGREE
SOLUTION TWO----
angle C also =180degree-68degree
=132degree
Q1 The problem becomes ambiguous when the triangle is a congruent as it is SSA and therefore, it may not be determined from the given conditions.
Q2In Triangle ABC, a=12cm, c=18cm & Angle A=30 deg. Find Angle C.
1st Solution:
12/sin30 = 18/sinC
sinC= 0.75
C= sin-1(0.75)
=48.6
2nd Solution:
180deg-48.6deg
=131.4 deg
Therefore, Angle C = 48.6deg or 131.4deg.
Note: deg means degrees
.: This question and solution demonstrates Ambiguous case when applying the SINE rule
Q1 The problem becomes ambiguous when the triangle is a congruent as it is SSA and therefore, it may not be determined from the given conditions.
Q2In Triangle ABC, a=12cm, c=18cm & Angle A=30 deg. Find Angle C.
1st Solution:
12/sin30 = 18/sinC
sinC= 0.75
C= sin-1(0.75)
=48.6
2nd Solution:
180deg-48.6deg
=131.4 deg
Therefore, Angle C = 48.6deg or 131.4deg.
Note: deg means degrees
.: This question and solution demonstrates Ambiguous case when applying the SINE rule
1.A problem becomes ambiguous when u apply SINE.When it is SSA( two sides and a no-n included angle) is not a condition for congruent triangles. Hence a triangle may not be uniquely determined the given conditions.
2.In triangle EFG,e=15cm,g=22cm,E=39degree.find angle G.
15/sin39deg=22/sin G
=0.923
=67.4deg or 180deg-67.4deg=112.6deg
angleG=67.4deg or 112.6deg.
1.A problem becomes ambiguous when u apply SINE.When it is SSA( two sides and a no-n included angle) is not a condition for congruent triangles. Hence a triangle may not be uniquely determined the given conditions.
2.In triangle ABC,a=17cm,c=23cm,A=37degree.find angle C.
17/sin37deg=23/sin C
=0.814
=54.5deg or 180deg-54.5deg=125.5deg
angleC=54.5deg or 125.5deg.
OH MY GOSH!!!!
mr.malek!!!! help!!!!
wo de tian ah!!!
don understand hor...die~~
-huan-
oppsss...
hmm...i tink u should put down de
graph? or fig.? hmm..
cos i cant understand through words la..i wan pic!! i wan fig.!!
-huan-
Q1: It is when SSA (two sides and a non-included angle) is not a condition for congruent triangles. Hence a triangle may not be uniquely determuned from the given conditions.
Q2: In △ABC, a=10cm, c=15cm and ∠A=40°.Find ∠C.
a/sin A = c/sin C
10/sin 40° = 15/sin C
sin C = 15 sin 40°/10
sin C = 0.964 (3 d.p.)
∠C = 74.6° or (180°- 74.6°)
= 74.6° or 105.4°
Q 1....
the problem become ambiguous when
b Sin B< a< b.....
Q 2....
in triangle ABC, a=8 , c=12 and < A=40degree. find < c!!!!!
solution!!!!
(8/sin40)=(12/sinC)
sin C= ((12(sin40)):8)
C = sin*(-1)X((12(sin40)):8)
c =74.6 (to 1 decplace)
or
c =105.4 (to 1 decplace)
*to the power of
ida ^^
Q1:the problem becomes ambiguous
when we given 2 sides and a non-
included angle of a triangle,
give us two possible
sollutions.
Q2:in ▲ABC,b=7cm,c=11cm
and < B=31deg.Find < C.
sollution:
7/sin31 = 11/sinC
sinC = 0.80934
< C = sin-1(0.80934)
< C = 54.0314
< C = 54.0 deg (1dcp)
or
< C = 180deg-54.0314deg
= 125.9686deg
= 126.0 deg (1dcp)
-frensky-
I dont understand the mean of question
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