Given 3 sides of a triangle, compute interior angles and area

[ai93]

Just attempted another exam question. Would you mind correcting me if I am wrong?
The question is
A triangle ABC has sides of length AB= 3.5 m, BC = 5.1 m and AC = 4.2m.

a) Calculate the size of the angle B and the size of the angle C, in degrees correct to 1 decimal place, in each case.

b) Calculate the area of triangle ABC.

My solution

\(\displaystyle a^{2}=b^{2}+c^{2}-2bcCOSA\)

\(\displaystyle \therefore COSA=\frac{b^{2}+c^{2}-a^{2}}{2bc}\)

\(\displaystyle COSA=\frac{4.2^{2}+3.5^{2}-5^{2}}{2\cdot4.2\cdot3.5}\)

A\(\displaystyle =COS^{-1}(0.132)\)
A= 82.4

Sine Rule

\(\displaystyle \frac{sin(82.4)}{5.1}=\frac{SinB}{4.2}\)

SinB=\(\displaystyle \frac{sin(82.4)\cdot4.2}{5.1}\)

B=Sin\(\displaystyle ^{-1}(0.816)

B=54.7\)

Now, 180-sin\(\displaystyle ^{-1}(0.816)=125.3\)
So angle B can equal 54.7 or 125.3

Angle C

180-(82.4+54.7)=42.9<180
or
180-(82.4+125.3)=-27.6 (which cannot be possible)
\(\displaystyle \therefore\)C = 42.9 degrees

and, 42.9+82.4+54.7=180

Question b)
Area=\(\displaystyle \frac{1}{2}bccosC\)

\(\displaystyle \frac{1}{2}\cdot5.1\cdot4.2 \cdot cos(42.9)\)=7.8m

Hope you understand that, I put what you told me into practice and hopefully solved this question correctly.

 

[MarkFL]

I have moved this post into a new thread...we ask that new questions not be tagged onto existing threads so that our threads don't become convoluted and hard to follow.

This is how I would work the problem:

First, we may use the Law of Cosines to find either angle B or C...I will find B first:

\(\displaystyle B=\arccos\left(\frac{3.5^2+5.1^2-4.2^2}{2\cdot3.5\cdot5.1}\right)=\arccos\left(\frac{1031}{1785}\right)\approx54.7^{\circ}\)

Next, use the Law of Sines to find angle C:

\(\displaystyle C=\arcsin\left(\frac{3.5\sin\left(\arccos\left(\dfrac{1031}{1785}\right)\right)}{4.2}\right)\approx42.9^{\circ}\)

There is no ambiguous case here, these are the only possible values.

Now, to find the area in square meters, we can use the formula:

\(\displaystyle A=\frac{1}{2}ab\sin(C)=\frac{1}{2}\cdot5.1\cdot4.2\cdot\frac{3.5\sin\left(\arccos\left(\dfrac{1031}{1785}\right)\right)}{4.2}=\frac{2}{25}\sqrt{8294}\approx7.3\)

 

[Prove It]

I have moved this post into a new thread...we ask that new questions not be tagged onto existing threads so that our threads don't become convoluted and hard to follow.

This is how I would work the problem:

First, we may use the Law of Cosines to find either angle B or C...I will find B first:

\(\displaystyle B=\arccos\left(\frac{3.5^2+5.1^2-4.2^2}{2\cdot3.5\cdot5.1}\right)=\arccos\left(\frac{1031}{1785}\right)\approx54.7^{\circ}\)

Next, use the Law of Sines to find angle C:

\(\displaystyle C=\arcsin\left(\frac{3.5\sin\left(\arccos\left(\dfrac{1031}{1785}\right)\right)}{4.2}\right)\approx42.9^{\circ}\)

There is no ambiguous case here, these are the only possible values.

Now, to find the area in square meters, we can use the formula:

\(\displaystyle A=\frac{1}{2}ab\sin(C)=\frac{1}{2}\cdot5.1\cdot4.2\cdot\frac{3.5\sin\left(\arccos\left(\dfrac{1031}{1785}\right)\right)}{4.2}=\frac{2}{25}\sqrt{8294}\approx7.3\)

The OP already has all three sides of the triangle, so Heron's Formula will be the most direct and exact way to get the area in this case...

 

[MarkFL]

The OP already has all three sides of the triangle, so Heron's Formula will be the most direct and exact way to get the area in this case...

Yep, that's another route one could take. :D

 

[ai93]

The OP already has all three sides of the triangle, so Heron's Formula will be the most direct and exact way to get the area in this case...

Gotten the angles correct, however got the area wrong.

I used this formula, \(\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)

Where s = \(\displaystyle \frac{a+b+c}{2}\)

\(\displaystyle \therefore s = \frac{5.1+4.2+3.5}{2}\)=6.4

\(\displaystyle \sqrt{6.4(6.4-5.1)(6.4-4.2)(6.4-3.5)}\)=7.28 \(\displaystyle \approx7.3m\)

 

[Prove It]

Gotten the angles correct, however got the area wrong.

I used this formula, \(\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)

Where s = \(\displaystyle \frac{a+b+c}{2}\)

\(\displaystyle \therefore s = \frac{5.1+4.2+3.5}{2}\)=6.4

\(\displaystyle \sqrt{6.4(6.4-5.1)(6.4-4.2)(6.4-3.5)}\)=7.28 \(\displaystyle \approx7.3m\)

Why do you think it's wrong? It's pretty much the same as what Mark got...

Also, like I said, because you can keep the answer just as a surd (without any mention of angles), if you keep it in its surd form you will get the EXACT answer, not just a decimal...

$\displaystyle \begin{align*} \sqrt{6.4 \left( 6.4 - 5.1 \right) \left( 6.4 -4.2 \right) \left( 6.4-3.5 \right) } &= \sqrt{ 53.0816} \\ &= \sqrt{\frac{33\,176}{625}} \\ &= \frac{2\,\sqrt{8294}}{25} \end{align*}$

This is the same exact answer Mark got...

 

[MarkFL]

The OP already has all three sides of the triangle, so Heron's Formula will be the most direct and exact way to get the area in this case...

You actually make a good point here...it is preferable to use a formula that only relies on given data, not results we compute from this data. I wanted to show the OP that he was using an incorrect formula for the area.