Quick question (1 Viewer)

Feynman · Oct 27, 2014

Hey, what's x?

seventhroot · Oct 27, 2014

come on m9

strong troll/10

Joshmosh2 · Nov 1, 2014

Find other angles by separating to individual triangles

InteGrand · Jan 12, 2015

Feynman said:
Hey, what's x?

sy37 said:
the diagram is not to scale therefore impossibru to tell

trick question nice one m8

The question definitely has an answer. The triangle ABC is a unique one in shape, since we know all its angles (it's an isosceles triangle with base angle 80°), and there is only one point E on BC such that ∠BAE = 70°, so point E is fixed with respect to ΔABC. Also, point D is fixed, as there is only one point on AC such that ∠ABD = 60°. Therefore, ∠AED is fixed, i.e. a single value for x exists. We find this value below.

Without loss of generality, let AB = 1. Using the fact that the angle sum of a triangle is 180°, we note that ∠AEB = 30° (from ΔAEB), ∠ADB = 40° (from ΔABD) and hence ∠BDE = 130° – x (from ΔBDE).

By the sine rule in ΔBDE,

$\frac{BE}{\sin (130 ^\circ - x)}=\frac{BD}{\sin (30^\circ + x)}$

$\Leftrightarrow \frac{BE}{\sin (50 ^\circ + x)}=\frac{BD}{\sin (30^\circ + x)}$ , since sin(130° – x) = sin[180° – (130° – x)] = sin(50° + x).

$\Leftrightarrow \frac{\sin (30^\circ + x)}{\sin (50 ^\circ + x)}=\frac{BD}{BE} \text{ }(\ast)$ .

By the sine rule in ΔABD,

$\frac{BD}{\sin 80 ^\circ}=\frac{AB}{\sin 40 ^\circ}$

$\Rightarrow BD = AB\cdot\frac{\sin 80 ^\circ}{\sin 40^\circ}=\frac{\sin 80 ^\circ}{\sin 40^\circ}$

$\Rightarrow BD=\frac{2\sin 40 ^\circ \cos 40^\circ}{\sin 40^\circ}=2\cos 40^\circ$ (by the double angle formula for sine).

By the sine rule in ΔBEA,

$\frac{BE}{\sin 70 ^\circ}=\frac{AB}{\sin 30 ^\circ}$

$\Rightarrow BE=\frac{\sin 70 ^\circ}{\sin 30 ^\circ} = 2\sin 70 ^\circ$ (as AB = 1 and cosec 30° = 2)

Substituting the values for BD and BE into Equation (*),

$\frac{\sin (30^\circ + x)}{\sin (50 ^\circ + x)}=\frac{BD}{BE}$

$=\frac{2\cos 40^\circ}{2\sin 70 ^\circ}$

$=\frac{\sin 50^\circ}{\sin 70 ^\circ}$ , as cos 40° = sin(90° – 40°) = sin 50°
i.e. $\frac{\sin (30^\circ + x)}{\sin (50 ^\circ + x)}= \frac{\sin 50^\circ}{\sin 70 ^\circ}$ .

It is clear that the LHS above becomes the RHS when x = 20°. If x = 20° is the only possible solution given the circumstances, the answer is x = 20°. We show that this is indeed the case.

Firstly, note that 0° < x < 70°. To see this, let D′ be the point on CB such that $DD^\prime\parallel AB$ .

Then by alternate angles in parallel lines DD′ and AB, ∠D′DB = ∠ABD = 60°. Then by angle sum of ΔBDE, we have

(x + 30°) + 20° + (60° + ∠EDD′) = 180° (as ∠BDE = 60° + ∠EDD′)
⇒ x + ∠EDD′ = 70°
⇒ 0° < x < 70°.

Now, consider $f(\theta):=\frac{\sin (30^\circ + \theta)}{\sin(50^\circ + \theta)}, \text{ for }\theta \in (0^\circ, 70^\circ)$ . The function is defined for all θ in the given domain.

Also,

$f^\prime(\theta)=\frac{\cos (30^\circ + \theta)\sin(50^\circ + \theta)-\sin(30^\circ + \theta)\cos(50^\circ + \theta)}{\sin^2(50^\circ + \theta)}$

$=\frac{\sin[(50^\circ + \theta)-(30^\circ + \theta)]}{\sin^2(50^\circ + \theta)}$

$=\frac{\sin 20^\circ}{\sin^2(50^\circ + \theta)}$

$>0 $ $\forall \theta \in (0^\circ, 70^\circ)$ .

Therefore, f(θ) is an increasing function (and hence one-to-one) for all θ in the given domain, so there can only be one solution to Equation (*), as 0° < x < 70°. As x = 20° is a solution, it is the only solution, and hence the answer to the original problem.

Quick question (1 Viewer)

Feynman

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seventhroot

gg no re

Joshmosh2

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InteGrand

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