The diagram shows a triangle ABC where AB = AC, BC = AD and Angle BAC = 20 Degrees. What is Angle ADB?

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The diagram shows a triangle ABC where AB = AC, BC = AD and Angle BAC = 20 Degrees. What is Angle ADB?

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Answer 1 · 2017-12-25T23:49:28

$∠ A D B = 150^{\circ}$ $/_ADB=150^@$ approximately.

Explanation:

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Let's let $∠ B D C = ∠ x$ $/_BDC= /_x$ and $∠ D B C = ∠ y$ $/_DBC= /_y$

Since triangle $A B C$ $ABC$ is an isosceles triangle then $∠ A C B = 80^{\circ}$ $/_ACB=80^@$

Then from triangle $B D C$ $BDC$ we have:

$\frac{B C}{sin x} = \frac{B D}{sin 80} = \frac{D C}{sin y}$ $(BC)/sinx=(BD)/sin80=(DC)/siny$

From triangle $A B C$ $ABC$ we have:

$\frac{A C}{sin 80} = \frac{B C}{sin 20}$ $(AC)/sin80=(BC)/sin20$

But $A C = A D + D C$ $AC=AD+DC$ and $B C = A D$ $BC=AD$ . Therefore:

$A C = B C + D C$ $AC=BC+DC$

$\frac{B C + D C}{sin 80} = \frac{B C}{sin 20}$ $(BC+DC)/sin80=(BC)/sin20$

$B C sin 20 + D C sin 20 = B C sin 80$ $BCsin20+DCsin20=BCsin80$

$D C sin 20 = B C (sin 80 - sin 20)$ $DCsin20=BC(sin80-sin20)$

$D C = \frac{B C (sin 80 - sin 20)}{sin 20}$ $DC=(BC(sin80-sin20))/sin20$

$\frac{B C}{sin x} = \frac{\frac{B C (sin 80 - sin 20)}{sin 20}}{sin y}$ $(BC)/sinx=((BC(sin80-sin20))/sin20)/siny$

$\frac{B C}{sin x} = \frac{B C (sin 80 - sin 20)}{sin y sin 20}$ $(BC)/sinx=(BC(sin80-sin20))/(sinysin20)$

$(cancel color(red)(BC))/sinx=(cancelcolor(red)(BC)(sin80-sin20))/(sinysin20)$

$sinysin20=sinx(sin80-sin20)$

From triangle $BCD$ , we know $x+y=100^@$

$y=100-x$

$sin(100-x)sin20=sinx(sin80-sin20)$

Using the trigonometric identity:

$sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta$

$(sin100cosx-cos100sinx)sin20=sinx(sin80-sin20)$

$sin20sin100cosx-sin20cos100sinx=sinx(sin80-sin20)$

$sin20sin100cosx=sinx(sin80-sin20+sin20cos100)$

Dividing both sides by $cosx$

$sin20sin100=tanx(sin80-sin20+sin20cos100)$

Solving for $tanx$ , we get:

$tanx=(sin20sin100)/(sin80-sin20+sin20cos100)$

Now, we can plug in the values of $sin$ and $cos$ for each angle and get a value for $tanx$ :

$tanx=((0.342)(0.9848))/((0.9848-0.342+(0.342)(-0.1736))$

$tanx=0.3368/0.5834=0.5773$

$x=arctan0.5773=29.9978^@$ or approximately $30^@$

$/_ADB=180-/_BDC=180-/_x=180-30=150^@$

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The diagram shows a triangle ABC where AB = AC, BC = AD and Angle BAC = 20 Degrees. What is Angle ADB?

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