Find the angles of a triangle ABC

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Find the angles of a triangle ABC

Postby triangle » Thu May 19, 2011 5:59 am

Find the angles of a triangle ABC if the relation between the sides is: AB:BC:CA = 5:6:7
triangle
 

Re: Find the angles of a triangle ABC

Postby Guest » Sun Jun 05, 2011 11:18 pm

[tex]50^\circ , 60^\circ , 70^\circ[/tex]
Guest
 

Re: Find the angles of a triangle ABC

Postby Guest » Mon Jun 06, 2011 3:08 am

Can you show me the proof, please?
Guest
 

Re: Find the angles of a triangle ABC

Postby Jessica » Thu Aug 23, 2012 10:08 am

Sum of three angles of a triangle = 180 degrees
Here the relation between AB:BC:CA = 5:6:7
Let us suppose that the angles are as below:
AB= 5x
BC=6x
CA=7x

Also, we already know that AB+BC+CA=180
5x+6x+7x = 180
18x = 180
x=10

So The angles are as below:
AB= 50 degrees
BC = 60 degrees
CA = 70 degrees.

(Proved) :D

Jessica
 
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Re: Find the angles of a triangle ABC

Postby AlanAnderson » Fri Dec 12, 2014 9:28 am

Yes this is the right way to solve the problem.

AlanAnderson
 

Re: Find the angles of a triangle ABC

Postby leesajohnson » Thu Nov 03, 2016 6:23 am

The Sum of angles of triangle is 180 degree
So the sides AB + BC + CA = 180
=> AB:BC:CA = 5:6:7
=> 5x + 6x + 7x = 180 ---- (i)
=> 18x = 180
=> x = 180/18
=> x = 10 ----- (ii)
By putting value of x into it

AB = 5x = 5(10) = 50 degree
Similarly BC = 60 degree
and CA = 70 degree
leesajohnson
 

Re: Find the angles of a triangle ABC

Postby Guest » Thu Nov 03, 2016 4:44 pm

All these posts are totally wrong. The question clearly states that 5:6:7 is the relationship of the sides, and asks for the values of the angles. To find the angles you should be applying the cosine rule.

R. Baber.
Guest
 

Re: Find the angles of a triangle ABC

Postby togiap » Wed Jun 21, 2017 6:38 am

I agree all these posts are totally wrong.To find the angles you should be applying the cosine rule.
AB = 5x; BC = 6x; CA = 7x
[tex]cos A = \frac{AB^{2}+AC^{2} - BC^{2}}{2.AB.AC}[/tex]

togiap
 
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Re: Find the angles of a triangle ABC

Postby Baltuilhe » Sat Dec 29, 2018 12:24 pm

Good afternoon!

There is another way to find the 3 angles, using a half-angle tangent formula, like this:
[tex]\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{r}{p-a}[/tex]

Where:
[tex]p=\dfrac{a+b+c}{2}\text{ and}\\r=\sqrt{\dfrac{(p-a)(p-b)(p-c)}{p}}[/tex]

So, to a 5:6:7 proportion, we calc p, r, and after that the 3 angles (or half-angles):
[tex]p=\dfrac{5+6+7}{2}=9\\ r=\sqrt{\dfrac{(9-5)(9-6)(9-7)}{9}}=\sqrt{\dfrac{4.3.2}{9}}=\dfrac{2\sqrt{6}}{3}\\\\ \tan\left(\dfrac{\alpha}{2}\right)=\dfrac{r}{p-a}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-5}=\dfrac{\sqrt{6}}{6}\Rightarrow \boxed{\alpha\approx 44^{\circ}\;24'}\\\\ \tan\left(\dfrac{\beta}{2}\right)=\dfrac{r}{p-b}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-6}=\dfrac{2\sqrt{6}}{9}\Rightarrow \boxed{\beta\approx 57^{\circ}\;6'}\\\\\ \tan\left(\dfrac{\gamma}{2}\right)=\dfrac{r}{p-c}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-7}=\dfrac{\sqrt{6}}{3}\Rightarrow \boxed{\gamma\approx 78^{\circ}\;30'}[/tex]

I hope I have helped! :)

Baltuilhe
 
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Re: Find the angles of a triangle ABC

Postby Baltuilhe » Tue Jan 01, 2019 6:04 pm

Good afternoon!

I don't know what happened to Latex, but, now i'm posting again (fixing the problem).
Can a tutor erase the last one? With errors in Latex? Thanks!

There is another way to find the 3 angles, using a half-angle tangent formula, like this:
[tex]\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{r}{p-a}\\\\\tan\left(\dfrac{\beta}{2}\right)=\dfrac{r}{p-b}\\\\\tan\left(\dfrac{\gamma}{2}\right)=\dfrac{r}{p-c}[/tex]

Where:
[tex]p=\dfrac{a+b+c}{2}\text{ and}[/tex]

[tex]r=\sqrt{\dfrac{(p-a)(p-b)(p-c)}{p}}[/tex]

So, to a 5:6:7 proportion, we calc p, r, and after that the 3 angles (or half-angles):
[tex]p=\dfrac{5+6+7}{2}=9[/tex]

[tex]r=\sqrt{\dfrac{(9-5)(9-6)(9-7)}{9}}=\sqrt{\dfrac{4.3.2}{9}}=\dfrac{2\sqrt{6}}{3}[/tex]

[tex]\tan\left(\dfrac{\alpha}{2}\right)=\dfrac{r}{p-a}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-5}=\dfrac{\sqrt{6}}{6}\Rightarrow \boxed{\alpha\approx 44^{\circ}\;24'}[/tex]

[tex]\tan\left(\dfrac{\beta}{2}\right)=\dfrac{r}{p-b}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-6}=\dfrac{2\sqrt{6}}{9}\Rightarrow \boxed{\beta\approx 57^{\circ}\;6'}[/tex]

[tex]\tan\left(\dfrac{\gamma}{2}\right)=\dfrac{r}{p-c}=\dfrac{\dfrac{2\sqrt{6}}{3}}{9-7}=\dfrac{\sqrt{6}}{3}\Rightarrow \boxed{\gamma\approx 78^{\circ}\;30'}[/tex]

[tex]\alpha+\beta+\gamma=180^{\circ}[/tex]

I hope I have helped! :)

Baltuilhe
 
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Re: Find the angles of a triangle ABC

Postby Guest » Thu Aug 06, 2020 9:01 pm

Please help to solve the question!!!
Find the indicated side in the spherical triangle:
a) B= 104 30.7' C= 62 52.1' a=56 6.4' to find c-?
b) A= 116 1.8' B= 103 17.6' C= 94 21.2' to find b-?

Many thanks in advance!
Guest
 

Re: Find the angles of a triangle ABC

Postby Maxut » Sat Oct 22, 2022 6:46 am

Jessica wrote:Sum of three angles of a triangle = 180 degrees
Here the relation between AB:BC:CA = 5:6:7
Let us suppose that the angles are as below:
AB= 5x
BC=6x
CA=7x

Also, we already know that AB+BC+CA=180
5x+6x+7x = 180
18x = 180
x=10

So The angles are as below:
AB= 50 degrees
BC = 60 degrees
CA = 70 degrees.

(Proved) :D


Thank you! You helped me a lot! I don't like doing math homework and I don't like studying this subject.

Maxut
 

Re: Find the angles of a triangle ABC

Postby Guest » Sun Oct 23, 2022 3:16 pm

Guest wrote:All these posts are totally wrong. The question clearly states that 5:6:7 is the relationship of the sides, and asks for the values of the angles. To find the angles you should be applying the cosine rule.

R. Baber.


Maybe you're right... I didn't think about the Pythagorean theorem.
Guest
 


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