# Is this question solvable using the law of sines and cosines?

1. Is it possible to (analytically) calculate the area of the following triangle using the rules of sines and cosines?
2. Is it possible to calculate it using only the rule of sines?

The data given is: $$BD$$ is a median to $$AC$$, $$\angle ABD = 50^\circ$$, $$\angle DBC = 18^\circ$$ and $$BD = 4.2~\text{cm}$$.

It seems to me that some data is missing.

Am I right?

• Doesn't look like anything is missing. I had to solve an equation numerically but I got BDC=about $138^\circ$ and then you can fill everything in from there. The median information was critical since that allows you to decouple the angles from the unknown lengths $AD$ and $DC$.
– Ian
Commented May 6, 2022 at 14:44
• This is for high-school students who cant solve numerical equations. The meaning is to give an analytical solution Commented May 6, 2022 at 15:09
• There is no need in numerical solution here, answer can be given as analytical expression containing input data. You didn't show that BD is median in picture, so I firstly decided that some information is missing. Now I see that information is enough to recover all sides and find area. Let $\angle ABD=\alpha$, $\angle CBD=\beta$, $BD=a$, $\angle BDA=\gamma$ (unknown), $AD=CD=b$ (unknown). Then use sine rule in $ABD$ and $CBD$. Then solve for $\gamma$ and $b$. Then $S=ab\sin\gamma$. Commented May 6, 2022 at 15:21
• @IvanKaznacheyeu Can you write out how you get $\gamma$ without numerics? I don't see how to do it, since you end up with an equation involving two sinusoids with disparate phases and you can't write it as $\sin(\dots)=\sin(\dots)$. Specifically I had $x=BDC$ and then $\frac{\sin(162^\circ-x)}{\sin(x-50^\circ)}=\frac{\sin(18^\circ)}{\sin(50^\circ)}$. I could see there being a shortcut to get specifically the area analytically however; I missed that this was what the question was asking for at first.
– Ian
Commented May 6, 2022 at 15:31
• One can solve this as trigonometrical equation. $\sin(162^\circ-x)\sin(50^\circ)=A\sin x+B\cos x$, the same for $\sin(x-50^\circ)\sin(18^\circ)$. Then divide by $\cos x$ and you can find $\tan x$. But solution of L.F. given in answer is much simpler. Commented May 6, 2022 at 15:34

Hint: extend $$BD$$ to $$E$$, such that $$DE = BD$$. You should be able to find enough information in the triangle $$\triangle BCE$$ to calculate its area.

Let $$x = AB$$ , $$y = BC$$, $$\beta = \angle BDA$$

Then

$$\text{Area} = \dfrac{1}{2} (4.2) ( x \sin(50^\circ) + y \sin (18^\circ) ) = \dfrac{1}{2} x y \sin(68^\circ)\hspace{15pt} (1)$$

And from the law of sines we have

$$\dfrac{ x }{\sin \beta } = \dfrac{AD}{\sin(50^\circ)}$$

$$\dfrac{y}{\sin \beta } = \dfrac{DC}{\sin(18)^\circ}$$

So by dividing these two equations we get

$$\dfrac{x}{y} = \dfrac{ \sin(18^\circ)}{ \sin(50^\circ) } \hspace{15pt} (2)$$

Equation $$(2)$$ implies that

$$y = \left(\dfrac{ \sin(50^\circ)}{ \sin(18^\circ) } \right) x \hspace{15pt} (3)$$

Substituting for $$y$$ from $$(3)$$ into $$(1)$$, and dividing by $$x$$ (because $$x$$ cannot be zero), results in

$$(4.2) ( 2 \sin(50^\circ) ) = x \dfrac{\sin(68^\circ) \sin(50^\circ) }{\sin(18^\circ)}$$

Hence,

$$x = \dfrac{ 8.4 \sin(18^\circ) }{\sin(68^\circ)}$$

and, consequently,

$$y = \dfrac{ 8.4 \sin(50^\circ) }{ \sin(68^\circ)}$$

And finally, the area is

$$\text{Area} = \dfrac{1}{2} x y \sin(68^\circ) = 35.28 \dfrac{ \sin(18^\circ) \sin(50^\circ) }{ \sin(68^\circ)} \approx 9.0074$$