Product of Side Lengths

Let $x$ and $y$ be the sides adjacent to the $60^\circ$ angle, and $z$ the opposite side. First we have Area = $\frac{1}{2} x\cdot y\cdot sin 60^\circ \Rightarrow \frac{90\sqrt{3}}{4} = \frac{x\cdot y}{2} \frac{\sqrt{3}}{2}\Rightarrow x\cdot y = 90$ .

Now, by the law of cosines we have: $z^2 = x^2 +y^2 - 2\cdot x\cdot y\cdot cos60^\circ\Rightarrow z^2 = x^2 +y^2 - x\cdot y$ .

We also have $x+y+z = 30\Rightarrow z^2 = (30 - x - y)^2$ $\Rightarrow z^2 = 900 + x^2 + y^2 - 60\cdot x - 60\cdot y + 2\cdot x\cdot y$ .

Using all three equations we arrive at $x + y = \frac{39}{2}\Rightarrow z = \frac{21}{2}\Rightarrow x\cdot y\cdot z = 90\cdot \frac{21}{2} = 945_\blacksquare$

$x,y,z$ are sides of $\triangle ABC$ Suppose : angle $(x,y)=60 ^\circ$ .

so $S=\frac{1}{2} xy\sin 60^\circ =\frac{90\sqrt{3}}{4}$ , hence $xy = 90$ .

By cosine formula, $z^2=x^2+y^2-2xycos60$ .
-> $(30-x-y)^2=x^2+y^2-90$ (cause $x+y+z=30 , xy=90$ )
-> $900-60(x+y)+2xy= -90$
-> $x+y=39/2$ (cause $xy= 90$ )
-> $z=21/2$
-> $xyz = 90 \times 21/2 =945$

[Latex edits]

We can assume that triangle ABC has AB = c, BC = a, CA = b and \angle BAC = 60 ^ \circ. [ABC] = \frac {1}{2}bc\sin \angle BAC = \frac {90\sqrt{2}}{4} Therefore bc = 90.

Then we apply the law of cosine: a^2 = b^2 + c^2 - 2bc\cos\angle BAC = b^2 + c^2 - 90 a^2 + a^2 = a^2 + b^2 + c^2 - 90 2a^2 + 90 = a^2 + b^2 + c^2

We also have a + b + c = 30 Then (a + b + c)^2 = 900 Or a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 900 Because a^2 + b^2 + c^2 = 2a^2 + 90 and bc = 90 Then a^2 + ab + ac = 315 Or a(a+b+c) = a*30 = 315 This leads to a = 21/2.

Because bc = 90, a = 21/2, then abc = 945.

Given a triangle with sides $a, b, c$ and $\alpha$ - the angle between $a$ and $b$ , we know that the area $A = \frac{ab \sin \alpha}{2}$ and that $c^2 = a^2 + b^2 - 2ab\cos \alpha$ (the later is known as the law of cosines). It is given that $\frac{ab \sin \alpha}{2} = \frac{90\sqrt{3}}{4}$ , also $a+b+c = 30$ and $\alpha = 60^\circ$ . It is well known that $\sin 60^\circ = \frac{\sqrt{3}}2$ and $\cos 60^\circ = \frac 1 2$ . From the area we get $ab = 90$ . From the law of cosines, $c^2 = a^2 + b^2 - ab$ . From the perimeter, $c^2 = (30 - a - b)^2 = 900 - 60(a+b) + a^2 + b^2 + 2ab$ . After putting the last two equations together, the squares cancel out and all that remains is $60(a+b) = 900 + 3ab = 1170 \implies a+b = \frac {39}{2}$ . Finally from perimeter, $c = 30 - (a+b) = \frac{21}2$ and $abc = 90 \cdot \frac{21}2 = 945$ .

[Note - Given $a+b = \frac {39}{2}, ab=90$ how would you solve for $a$ and $b$ ? - Calvin]

Recognize a random scalene triangle can be broken into two right triangles by drawing the altitude from any given vertex. Given that one angle is $60 ^ \circ$ we have one 30-60-90 triangle and the key to the problem.

Let the sides of our 30-60-90 triangle be $a$ , $2a$ , and $\sqrt{3}a$ respectively (the altitude is $\sqrt{3}a$ ). Concerning the side of our scalene with length $a$ let the remainder be $b$ . According to the Pythagorean theorem, the remaining side of the scalene has length $\sqrt{b^2+3a^2}$ . We can now describe the perimeter and area of our scalene triangle in terms of $a$ and $b$ ; with two unknowns and two equations we can now solve.

For the Area:

$\begin{array}{ r l } \text{Area} & = \frac{1}{2}a*\sqrt{3}a + \frac{1}{2}b*\sqrt{3}a \\ \frac{90\sqrt{3}}{4} & = \frac{1}{2}a\cdot \sqrt{3}a + \frac{1}{2}b\cdot \sqrt{3}a \\ \frac{90\sqrt{3}}{2} & = a*\sqrt{3}a + b\cdot \sqrt{3}a \\ \frac{90}{2} & = a\cdot a + b\cdot a \\ 45 & = a^2 + ab \\ b & = \frac{45 - a^2}{a} \\ \end{array}$

For the Perimeter: $\begin{array}{ r l } \text{Perimeter} & = 2a + a + b + \sqrt{b^2+3a^2} \\ 30 & = 3a + b + \sqrt{b^2+3a^2} \\ 30 - 3a - b & = \sqrt{b^2+3a^2} \\ (30 - 3a - b)^2 & = b^2+3a^2 \\ 900-180a-60b+9a^2+6ab+b^2 & = b^2+3a^2 \\ 900 -60b+6ab& = -6a^2+180a\\ \end{array}$

Substituting in the above: $\begin{array}{ r l } 900 -60(\frac{45-a^2}{a})+6a(\frac{45-a^2}{a}) & = -6a^2+180a\\ 900 -\frac{2700}{a}+60a+270-6a^2 & = -6a^2+180a\\ 1170 & = 120a + \frac{2700}{a}\\ 1170a & = 120a^2 + 2700\\ 0 & = 4a^2 -39a +90\\ 0 & = (2a-7.5)(2a-12)\\ \end{array}$

Potential solutions: $a=6,7.5$

Plugging $a=6$ into $b = \frac{45 - a^2}{a}$ we get $b=1.5$ . Which we plug into our perimeter equation $2a + a + b + \sqrt{b^2+3a^2}$ to verify is 30.

Checking the other solution $a=7.5$ , $b$ becomes negative.

We have but to solve the original question, what is the product of the lengths of the three sides:

The first side is $a + b = 6 + 1.5 = 7.5$ .

The second side is $2a=2 \cdot 6 = 12$ .

The third side is $\sqrt{b^2+3a^2} =\sqrt{1.5^2+3 \cdot 6^2} = 10.5$

So the solution is $7.5 \cdot 12 \cdot 10.5 = 945$

Q.E.D.

WLOG, let two sides that make up the 60 degree angle be be $BC=a$ and $AC=b$. Then by the given, $\frac{1}{2}ab\sin C=\frac{1}{2}ab\sin 60^\circ=\frac{90\sqrt{3}}{4}$ so $ab=90$ and thus $b=90/a$.

Now using the Law of Cosines, we find that the remaining side has length $$\sqrt{a^2+8100/a^2-90}.$$ However, the perimeter is 30 so it can also be expressed as $$30-a-90/a.$$ Setting these expression equal, we find that $$a={15/2, 12}$$ which are $a$ and $b$. Then $c=\frac{21}{2}$ so the product of the side lengths is $$12\left(\frac{15}{2}\right)\left(\frac{21}{2}\right)=945.$$

Let AB=c,BC=a and CA=b $$[ABC]=90\sqrt{3}/4$$ or, $$\frac {1}{2} \times AB \times AC \times \sqrt{3}/2=90\sqrt{3}/4$$ $$\Rightarrow b \times c=90$$ Again $$[ABC]=\sqrt {s} \times \sqrt {s-a} \times \sqrt {s-b} \times \sqrt {s-c}$$ where s=semiperimeter=15 $$\Rightarrow 15(15-a)(15-b)(15-c)=24300$$ $$(15-a)[225-15(b+c)+90]=\frac {1620}{16}$$ $$(15-a)[315-15(30-a)]=\frac {1620}{16}$$ Hence we get,$$4a^{2}-96a+567=0$$ Solving the eq. we get $$a=\frac {108}{8},\frac {84}{8}$$ $$\Rightarrow a \times b \times c=945$$

First, since one of the angles in this triangle is 60° (let’s call this angle C) we know that sin C = √3/2 and cos C= 1/2 .

The problems states that Area = (90√3)/4. Using the equation that Area = ½ ab sinC (where sides a and b are adjacent to angle A) ½ ab √3/2 = (90√3)/4 so the product ab = 90

Using a manipulation of the cosine rule, we know that cos⁡C=(a^2+b^2-c^2)/( 2ab). Substituting in cos C= 1/2, and ab=90, we get that: 1/2⁡=(a^2+b^2-c^2)/( 2(90)) or that 90 = a^2+b^2-c^2.

Using 90=a^2+b^2-c^2 and completing the square we obtain: 90=(a+b)^2-2ab-c^2 Since the perimeter is 30,we know a + b + c = 30, or a + b = 30 – c. Using this eqation with ab = 90, and substituting we find that:

90=(30-c)^2-2(90)-c^2

90=900-60c+c^2-180-c^2

90=900-60c-180

90=720-60c

So, c = 11.25.

Therefore abc = 90 × 11.25 = 945.

using hero's formula ; s(s-a)(s-b)(s-c)= A*A s = (a+b+c)/2=15 area= ab sin 60/2 -- ab=90; replace c as 30-(a+b); solve for a+b.. we get a+b = 39/2 -- c = 21/2-- abc = 945

Let AC be b,AB be c and bC be a Assuming ABC is 60 degree. Area of triangle =0.5acsin60 ac=90 The are many possible values of a and c, below are some of them 4 and 22.5 5 and 18 6 and 15 9 and 10 12 and 7.5 Using cosine rule,b^2=c^2+a^2-2accos60 We found that the values of a and c is 12 and 7.5. Hence b is 30-12-7.5= 10.5 Product of the sides of ABC =abc=12x7.5x10.5=945

let,angle B=60, a,b,c be the arms of the triangle ABC perimeter=a+b+c=30; area=0.5 a c sin(B); by writing the value of area,we get a c=90; we also know,area=under root(s(s-a)(s-b)(s-c)),where s=(a+b+c)/2 by solving,we get b=(21/2) or (27/2) but, if b=(27/2), a b c>999 so,according to the given condition, a b c=945. when b=(21/2)(i.e ans<999)

Without loss of generality, let $\angle BAC = 60^\circ$ . Let $[ABC]$ denote the area of triangle $ABC$ and let $CB=a, AC=b$ and $AB=c$ .

Thus $[ABC] = \frac {1}{2} bc \sin 60^\circ = \frac{\sqrt{3}}{4} bc = \frac {90 \sqrt3}{4} \Rightarrow bc = 90$ , and $a + b + c = 30$ . From the law of cosines we have $a^2=b^2+c^2-bc = (b+c)^2 - 3bc$ .

Substituting $bc = 90$ and $b+c = 30-a$ into the last equation gives $a^2 = (30-a)^2 - 3 \times 90 \Rightarrow a = \frac{21}{2}$ . Thus $b+c = \frac{39}{2}$ .

Substituting $b = \frac{39}{2}-c$ into $bc = 90$ , we get $\left(2c-15\right)\left(c-12\right) = 0$ which gives $b = \frac{15}{2}, c= 12$ or $b = 12, c = \frac{15}{2}$ .

Hence $abc = \frac{21}{2} \times \frac{15}{2} \times 12 = 945$ .

Two step solution is: Use formula for area of triangle $\frac{1}{2}ac sin(60)$ , and the other one is the half angle formulae which says that $tan (60/2)=\frac{Area of triangle}{s(s-b)}$ where s is semi-perimeter. And I think this also avoids the possibility of other solution.

Let the angle 60^\circ is between side a and b. Area of triangle ABC is $\frac{90\sqrt{3}}{4}=\frac{1}{2}ab\sin 60^\circ=\frac{ab\sqrt{3}}{4} \Rightarrow ab=90$ . Applying Heron's formula, we have $\sqrt{s(s-a)(s-b)(s-c)}=\frac{90\sqrt{3}}{4}=\sqrt{\frac{24300}{16}}$ $\Rightarrow s(s-a)(s-b)(s-c)=\frac{24300}{16}$ , where s is half of the perimeter, which is equal to 15. a,b and c are the length of sides of the triangle. So, $\frac{24300}{16}=50625-15abc+225(ab+bc+ca)-3375(a+b+c)$ $=50625-15(90)c+225[90+c(30-c)]-3375(30)$ $=-225c^2+5400c-30375$ . Applying quadratic formula on $12c^2-288c+1701=0$ , we have $c=10.5$ or $c=13.5$ . So, there are actually 2 answer, which are $abc=90\cdot 13.5 =1215$ or $abc=90\cdot 10.5=945$ .

Let the side lengths of the triangle be a,b, and c, and the angle of 60 be between sides a and b. Notice that by the Sine Area Formula, we have:

1/2 (a)(b)sin(60) = 90sqrt(3)/4 ab = 90

Also, by the Law of Cosines, we have:

c^2 = a^2 + b^2 - 2ab cos(60) = a^2 + b^2 - ab = (a+b)^2 - 3ab

We now make the substitution x = a+b. Thus, c = 30 - (a + b) = 30 - k. We substitute this into the above equation from the Law of Cosines:

(30 - k)^2 = k^2 -3(90) k^2 - 60k + 900 = k^2 - 270 k = 1170/60 = 39/2

Thus, c = 21/2, so the answer is abc = (90)(21/2) = 945.

(Note: The Preview did not render my LaTeX properly, so I didn't bother TeXing it.)

by using various formulas of trignometry we can find.

area=90.root3 /4=45.root3 /2. 1/2 product of two sides sine of angle between them=area of a triangle. by this,product of two sides=90. by using another area fomula,we can find that the third side is 10.5. thus product=90*10.5=945.

the triangle has sides a, b and c, one of its angles is 60, I will assume angle BCA = 60 (angle C) since everything is unknown it is possible.

perimeter formula : a+b+c= 30............... 1

cos rule : c^2= a^2+b^2-2ab cos 60...............2

area formula : 1/2 ab sin 60= 90√3/4.........................3

From 1,

c= 30-a-b

c^2= 900-60(a+b)+(a+b)^2

c^2= 900-60(a+b) + a^2+2ab+b^2

From 3,

ab = 90

Subs 3 to 1,

c^2= a^2 +b^2 - 60(a+b) +900 + 2x90

  = a^2 +b^2 - 60(a+b) +1080

subs 3 to 2,

c^2= a^2 + b^2 -90

compare equation for c^2,

a^2 + b^2 - 90 = a^2 +b^2 - 60(a+b) +1080

-90 = - 60(a+b) +1080

-60(a+b)= -1170

a+b= 19.5.................4

Subs 4 to 1

19.5+c=30

c=10.5

abc =90x 10.5

    =945

Let $\angle C = 60 ^ \circ$ .

[ABC] = $\frac12ab*\sin\angle C$ = $\frac{90\sqrt{3}}{4}$ . Then, ab = 90.

By the Law of Cosines, $c^2 = a^2 + b^2 - 2ab\cos\angle C$ . Noting that $c = 30 - a - b$ , then substituting for ab and $\cos\angle C$ , we get $(30 - a - b)^2 = a^2 + b^2 - 90$ , which expands and simplifies to give $495 + ab = 30 (a + b)$ .

Since ab = 90 and a + b = 30 - c, we rearrange to get $30c = 315$ , or c = 19.5. The desired answer equals abc, which is $90 * 19.5 = 585$ .