Engineering Mechanics - General Principles - Discussion
Discussion Forum : General Principles - General Questions (Q.No. 6)
6.
Determine the angles
and 
and the length of side AB of the triangle. Note that there are two possible answers to this question and we have provided only one of them as an answer. = 46.7°, = 93.3° d = 9.22 = 50.0°, = 90.0° d = 9.14 = 40.0°, = 100.0° d = 9.22 = 48.6°, = 91.4°, d = 9.33Discussion:
17 comments Page 1 of 2.
Preethi said:
1 decade ago
sin 40'= 6/ab
AB= 6/SIN 40
AB= 9.33
option d
AB= 6/SIN 40
AB= 9.33
option d
Hemu Singh said:
1 decade ago
If the sides of the triangle are a=6, b=7,and AB=d (let)
then according to cosine triangle formula
COS(A)= (b^2 + c^2 - a^2)/2bc
Where A=40'
d = 9.33 & 1.39
So ans. is option d....
then according to cosine triangle formula
COS(A)= (b^2 + c^2 - a^2)/2bc
Where A=40'
d = 9.33 & 1.39
So ans. is option d....
Hasi said:
1 decade ago
It would be better to give a small description for angles also...
Kumar said:
1 decade ago
Hai Preethi you said answer for this problem is only applicable for right angled triangle, but it is not a right angled triangle.
Nitish said:
1 decade ago
Using sine formula,
a/sin A = b/sin B
6/sin 40 = 7/sin CBA
a/sin A = b/sin B
6/sin 40 = 7/sin CBA
Damodaran said:
1 decade ago
I'm confused please let me clear.
Aviral Singh said:
1 decade ago
In triangle ABC, taking AC as the base of the triangle
i.e. 7 units
and height BC i.e. 6 units
Then, Sin40=6/AB
Hence, AB=6/Sin40
Therefore, AB=9.33
Now using sine formula 7/sinCBA=6/sin40
Therefore SinCBA=0.7499
so Angle CBA=48.58
Now AngleACB,, ACB+40+48.58=180
Angle ACB=91.42
ANSWER: Option D
i.e. 7 units
and height BC i.e. 6 units
Then, Sin40=6/AB
Hence, AB=6/Sin40
Therefore, AB=9.33
Now using sine formula 7/sinCBA=6/sin40
Therefore SinCBA=0.7499
so Angle CBA=48.58
Now AngleACB,, ACB+40+48.58=180
Angle ACB=91.42
ANSWER: Option D
Kcrkr said:
1 decade ago
Please let me know why you are using Sin theta, I think sin is using against for length / hypotenus, we get for cos theta also because we know base of triangle. Then answer is 9.22.
PRAMOD said:
1 decade ago
COS40=7/AB
AB=7/COS40
AB=9.13
AB=7/COS40
AB=9.13
Sujit kumar singh said:
1 decade ago
See you are right but sin(theta) is also used for side and angle relation.
Like formula is a/sin(alpha)=b/sin(bita)=c/sin(gama).
Where a,b,c are the sides of triangle and alpha, bita,gama are the angle infront of side of triangle like a, b, c.
So we can solve as:
7/sin(fai)=6/sin(40)
so sin(fai)=0.7499
and fai=48.58 degree.
Also theta=180-(40+48.58)=91.42 degree.
And using same formula as above we can find length AB as
6/sin(40)=AB/sin(91.42)
so AB=9.33
Like formula is a/sin(alpha)=b/sin(bita)=c/sin(gama).
Where a,b,c are the sides of triangle and alpha, bita,gama are the angle infront of side of triangle like a, b, c.
So we can solve as:
7/sin(fai)=6/sin(40)
so sin(fai)=0.7499
and fai=48.58 degree.
Also theta=180-(40+48.58)=91.42 degree.
And using same formula as above we can find length AB as
6/sin(40)=AB/sin(91.42)
so AB=9.33
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