Question

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

Answer 1

`/_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^@`