Find the Missing Angle in Triangle ABC: Use Arithmetic Sequence and Sine Rule

[Physiona]

QUESTION:

The angles in triangle ABC form an increasing arithmetic sequence.

The ratio of angles A:B:C can be written in the form 185:370:555 respectively.

You are told that the total area of the triangle is 9

Length BC is

Given the area of triangle ABC, work out the value of

Give your answer in the form

where

and

are integers.

ATTEMPT:
I have decided to figure out a multiplier for all the angles provided, and labelled it as y
y(185+370+555)= 180
1110y = 180
y= 6/37
I am then planning on using the sine rule however, I don't have set values to use only angles. I'm not sure on whether or not my attempt is correct. Any guidance to helping? Thank you!

 

[Mark44]

QUESTION:
View attachment 222906
View attachment 222907View attachment 222908

The angles in triangle ABC form an increasing arithmetic sequence.

The ratio of angles A:B:C can be written in the form 185:370:555 respectively.

You are told that the total area of the triangle is 9View attachment 222909

Length BC is View attachment 222910
Given the area of triangle ABC, work out the value of View attachment 222911
Give your answer in the form View attachment 222912 where View attachment 222913 and View attachment 222914 are integers.

ATTEMPT:
I have decided to figure out a multiplier for all the angles provided, and labelled it as y
y(185+370+555)= 180
1110y = 180
y= 6/37
I am then planning on using the sine rule however, I don't have set values to use only angles. I'm not sure on whether or not my attempt is correct. Any guidance to helping? Thank you!

The angles form an increasing arithmetic sequence. How have you used that information?

 

[Physiona]

The angles form an increasing arithmetic sequence. How have you used that information?

Erm, I think I've used it in this way.
After I have found out the multiplier of 6/37, I multiplied each part of the ratio by it.
185*6/37= 30
370*6/37= 60
555*6/37= 90
This eventually forms an arithmetic sequence right? 30, 60, 90. (Increasing by 30 each time)
I have used to sine rule to then find the length of AC.
a/sin a = b/sin b
x/sin(30) = AC/sin(60)
Rearranging for AC = is root3/4 x = AC
I've then used the area of a triangle to figure out x
A = 1/2ab*SinC
9= 1/2*1/2*root3/4x*x*sin(30)
18= 1/2*root3/4x²
36=root3/4x²
Rearranging it for x= root 48 over root 3.

 

[Mark44]

Erm, I think I've used it in this way.
After I have found out the multiplier of 6/37, I multiplied each part of the ratio by it.
185*6/37= 30
370*6/37= 60
555*6/37= 90

I got the same numbers, but in a more straightforward way, I think.
Let a = the smallest angle. The the three angles are a, a + h, a + 2h. Since (a + h)/a = 370/185, then 185(a + h) = 370a, or simplified, a + h = 2a, from which we get h = a. This means that the angles are a, 2a, and 3a, from which we get a = 30° and the other two angles are 60° and 90°.

Your drawing is not very helpful. Is A the angle at the left and C the angle on the right? And in your work below, which side is x?

This eventually forms an arithmetic sequence right? 30, 60, 90. (Increasing by 30 each time)
I have used to sine rule to then find the length of AC.

Since you have a 30-60-90 right triangle, it's much simpler to note that if x is the length of the short leg, then 2x is the length of the hypotenuse, and the other leg is $\sqrt 3 x$.

a/sin a = b/sin b
x/sin(30) = AC/sin(60)
Rearranging for AC = is root3/4 x = AC
I've then used the area of a triangle to figure out x
A = 1/2ab*SinC
9= 1/2*1/2*root3/4x*x*sin(30)
18= 1/2*root3/4x²
36=root3/4x²
Rearranging it for x= root 48 over root 3.

You have $\frac{\sqrt{48}}{\sqrt{3}} = \sqrt{16} = 4$. I get a different result, which is in the form of $\sqrt{a \sqrt b}$, which produces the right area and works in the Pythagorean Law.

 

[Physiona]

I got the same numbers, but in a more straightforward way, I think.
Let a = the smallest angle. The the three angles are a, a + h, a + 2h. Since (a + h)/a = 370/185, then 185(a + h) = 370a, or simplified, a + h = 2a, from which we get h = a. This means that the angles are a, 2a, and 3a, from which we get a = 30° and the other two angles are 60° and 90°.

Your drawing is not very helpful. Is A the angle at the left and C the angle on the right? And in your work below, which side is x?

Since you have a 30-60-90 right triangle, it's much simpler to note that if x is the length of the short leg, then 2x is the length of the hypotenuse, and the other leg is $\sqrt 3 x$.

You have $\frac{\sqrt{48}}{\sqrt{3}} = \sqrt{16} = 4$. I get a different result, which is in the form of $\sqrt{a + \sqrt b}$, which produces the right area and works in the Pythagorean Law.

Basically the diagram didn't upload right, so it was difficult to adjust it on here. The angle on the bottom left is A, on the top right is B, and on the bottom right is C (which is a right angle). I'm not sure how you have got $\sqrt 3 x$.
Can you please explain the method fully with what you have as answer? This is the method my teacher assumed I should follow.

 

[Physiona]

What does the $h$ represent in the equation used? Is it some constant?

 

[Mark44]

Basically the diagram didn't upload right, so it was difficult to adjust it on here. The angle on the bottom left is A, on the top right is B, and on the bottom right is C (which is a right angle). I'm not sure how you have got $\sqrt 3 x$.

If x is the length of the leg opposite the 30° angle, then the hypotenuse length is 2x, and the other leg length is $\sqrt 3 x$. From the Law of Pythagoras, we have $x^2 + (\sqrt 3 x)^2 = (2x)^2$.

What does the h represent in the equation used? Is it some constant?

Yes, it's the constant in the arithmetic sequence. If the sequence starts with a, the next term will be a + h, and the next a + 2h, and so on.

Knowing that the two legs of the triangle are $x$ and $\sqrt 3 x$, substitute them into your formula for the given area to find x, which is not equal to 4, as you had in your earlier post.

 

[Physiona]

If x is the length of the leg opposite the 30° angle, then the hypotenuse length is 2x, and the other leg length is $\sqrt 3 x$. From the Law of Pythagoras, we have $x^2 + (\sqrt 3 x)^2 = (2x)^2$.

Yes, it's the constant in the arithmetic sequence. If the sequence starts with a, the next term will be a + h, and the next a + 2h, and so on.

Knowing that the two legs of the triangle are $x$ and $\sqrt 3 x$, substitute them into your formula for the given area to find x, which is not equal to 4, as you had in your earlier post.

Right okay, I'll have an attempt.

 

[Physiona]

I'm attempting to use the formula of $1/2abSinC$.
I've got all the lengths of the triangle as $\sqrt 3 x$ and $x$ and $2x$. The using of the area of a triangle, will that lead me to finding a solution for x? I did it, but I didn't manage to get a correct solution in the form of $\sqrt{a + \sqrt b}$

If x is the length of the leg opposite the 30° angle, then the hypotenuse length is 2x, and the other leg length is $\sqrt 3 x$. From the Law of Pythagoras, we have $x^2 + (\sqrt 3 x)^2 = (2x)^2$.

Yes, it's the constant in the arithmetic sequence. If the sequence starts with a, the next term will be a + h, and the next a + 2h, and so on.

Knowing that the two legs of the triangle are $x$ and $\sqrt 3 x$, substitute them into your formula for the given area to find x, which is not equal to 4, as you had in your earlier post.

 

[Physiona]

I'm still struggling! I come to a final conclusion in finding $x$, but my calculation is misinterpreted somehow.
I get $\sqrt 3x$*$x$\2 $= 9$
I used the area of a triangle, $b*h/2$.

If x is the length of the leg opposite the 30° angle, then the hypotenuse length is 2x, and the other leg length is $\sqrt 3 x$. From the Law of Pythagoras, we have $x^2 + (\sqrt 3 x)^2 = (2x)^2$.

Yes, it's the constant in the arithmetic sequence. If the sequence starts with a, the next term will be a + h, and the next a + 2h, and so on.

Knowing that the two legs of the triangle are $x$ and $\sqrt 3 x$, substitute them into your formula for the given area to find x, which is not equal to 4, as you had in your earlier post.

 

[Mark44]

I'm still struggling! I come to a final conclusion in finding $x$, but my calculation is misinterpreted somehow.
I get $\sqrt 3x$*$x$\2 $= 9$

Now solve for x. You will need to do some algebraic manipulation to get it into the form $x = \sqrt{a \sqrt b}$.

I used the area of a triangle, $b*h/2$.

 

[Charles Link]

@Mark44 We were unaware of the duplication, but the OP was able to work to the answer on the other thread, so this one could be closed as well.

 

[Mark44]

@Mark44 We were unaware of the duplication, but the OP was able to work to the answer on the other thread, so this one could be closed as well.

The other thread, which was started after this one, was closed because it was a duplicate of this thread, not because the OP had found a solution.