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

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1 Answer
Dec 25, 2017

/_ADB=150^@ approximately.

Explanation:

.

Let's let /_BDC= /_x and /_DBC= /_y

Since triangle ABC is an isosceles triangle then /_ACB=80^@

Then from triangle BDC we have:

(BC)/sinx=(BD)/sin80=(DC)/siny

From triangle ABC we have:

(AC)/sin80=(BC)/sin20

But AC=AD+DC and BC=AD. Therefore:

AC=BC+DC

(BC+DC)/sin80=(BC)/sin20

BCsin20+DCsin20=BCsin80

DCsin20=BC(sin80-sin20)

DC=(BC(sin80-sin20))/sin20

(BC)/sinx=((BC(sin80-sin20))/sin20)/siny

(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^@