Find unknown side in triangle

Question

I encountered this problem which I am not sure how to solve.

Length of two sides are given as 250 ft and 60 ft.These sides are adjacent to each other.Measurement of one of the angles of triangle is given as 50 degrees. The angle is enclosed between side having length 60 ft and the side whose length needs to be found.

Below is the pic of problem.

Find x in this image

Answer 1

hint

Let $ a,b,c $ be respectively the opposite sides of the angles $ A,B,C $ of the triangle.

then we have an interesting identity

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

Answer 2

Let $|BC|=a=60$, $|AC|=b=250$, $|AB|=x$, $\angle ABC=\beta=50^\circ$.

Draw a perpendicular $CD\perp AB$. Then

\begin{align} |AB|=x&=|AD|+|BD|=x_1+x_2 . \end{align}

\begin{align} \triangle CDB:\quad x_2&=a\cos\beta ,\\ |CD|=h&=a\sin\beta ,\\ \triangle CAD:\quad x_1&= \sqrt{b^2-h^2} = \sqrt{b^2-a^2\sin^2\beta} ,\\ x=x_2+x_1 &= a\cos\beta +\sqrt{b^2-a^2\sin^2\beta} \\ &= 60\cos50^\circ +\sqrt{250^2-60^2\sin^2 50^\circ} \\ &\approx 284.3 . \end{align}

Answer 3

Easiest is cosine rule, as you need one side and its opposite angle to use the sine law easily.

Cosine rule is not entirely trivial here, it leads to a quadratic, but at least it's smooth sailing.

The choice of what goes on the $\mathrm{LHS}$ is dependent on what angle you're given. Here the angle is $50^{\circ}$ so $250$ (opposite the given angle) goes on the $\mathrm{LHS}$.

$250^2 = x^2 + 60^2 - 2(x)(60)\cos 50^{\circ}$

$x^2 - 120\cos 50^{\circ} x -58900 = 0$

Solving that (using quadratic formula) gives you:

$\displaystyle x = \frac{120\cos 50^{\circ} \pm \sqrt{(120\cos 50^{\circ})^2 - 4(-58900)}}{2}$

$x \approx 284.31$ or $x \approx -207.17$, and it's clear the negative root must be rejected.

So the answer is $x \approx 284.31$.