Question

My book doesn't have any examples regarding this kind of problem; could someone please help me out?

The area of a triangle is 75 square inches.
Find the length of the side included between A = 35° and C = 95°.

Given:
12 years ago

Answer 1

Are you given any other information, such as the length of any side?

Answer 2

No :/

Answer 3

I think I figured out a way to solve it.
Use the law of sines to find an expression in terms of a for the lengths of b and c.
Then use the formulas you have to write an expression for the area and set it equal to 75.
Then solve for a.

Answer 4

There aren't any equations that can be solved without at least one known side.

Answer 5

Wait. I gave you the wrong side above. You are looking for AC which is the side labeled b in the figure.

Use the law of sines to find an expression in terms of b for the lengths of a and c. Then use the formulas you have to write an expression for the area and set it equal to 75. Then solve for b.

Answer 6

You will have an equation with only one variable, b.
The right side will have the area you were given, 75.

Answer 7

Are you familiar with the law of sines?

Answer 8

Yes.
\[\frac{ a }{ \sin A } = \frac{ b }{ \sin B } = \frac{ c }{ \sin C }\] or
\[\frac{ \sin A }{ a } = \frac{ \sin B }{ b } = \frac{ \sin C }{ c }\]

Answer 9

So \[b = \frac{ c \sin 50 }{ \sin 95 } = 0.76897061\] and \[\frac{ 75 }{ 0.76897061 } = 97.53298654\] ?

Answer 10

Great.
Now use the first two fractions and find an expression for a in terms of b.
Then use the last two fractions and find an expression for c in terms of b.

Answer 11

The "Great" above refers to the fact you know the law of sines.

Answer 12

Keep b as b. That is what we are looking for.

Find a and c in terms of b.

Answer 13

\[a = \frac{ b \sin 35 }{ \sin 50 } \]
\[c = \frac{ b \sin 95 }{ \sin 50 }\]

Answer 14

Excellent. I got the same. See below.

\( \dfrac{a}{\sin A} =\dfrac{b}{\sin B} = \dfrac{c}{\sin C} \)

\( \dfrac{a}{\sin 35^o} =\dfrac{b}{\sin 50^o} = \dfrac{c}{\sin 95^o} \)

\(a = \dfrac{b \sin 35^o}{\sin 50^o} \) and \(c = \dfrac{b \sin 95^o}{\sin 50^o} \)

\(a = 0.74875b\) and \(c = 1.30044b\)

Answer 15

Now you can find s for Heron's formula.

Answer 16

\(s = \dfrac{1}{2}(a+b+c) \)

\(s = \dfrac{0.74875b + b + 1.30044b}{2} \)

\(s = \dfrac{3.04919b}{2} \)

\(s = 1.5245995b\)

Answer 17

\(A = \sqrt{s(s-a)(s-b)(s-c)} \)

Now replace s, a, and c with the expressions we have above in terms of b.
Then set it equal to 75.
Then simplify the inside of the root since everything is in terms of b.
Then square both sides and solve for b.

Answer 18

\[75 = \sqrt{ 1.5245995b (1.5245995b-0.74875b)(1.5245995b-b)(1.5245995b-1.30044b)}\]
\[= \sqrt{0.13909717b^4}\]
\[0.13909717b^4=5625\] (because 75^2 = 5625)
\[b^4 =\frac{ 5625 }{ 0.13909717 } = 40439.35617094\]
\[\sqrt[4]{40439.35617094} = 14.18081065\]

Answer 19

That's what I got.
Great job!

Answer 20

Yay! Thanks so much for your help. I couldn't have done it without you! :)

Answer 21

You're welcome. You did a great job. You just needed some help to get going!