Question

In triangle ABC, how do you solve the right triangle given Side AB=17 Side BC=22?

Answer 1

See the solution below

Explanation:

In right `\triangle ABC`, let the legs be `AB=17` & `BC=22`

Using Pythagorean theorem, in given right triangle the hypotenuse `AC` is given as

`AC=\sqrt{AB^2+BC^2}`

`=\sqrt{17^2+22^2}`

`=\sqrt773`

`=27.803`

Now, using sine formula in right triangle to find the angle `A` as follows

`\sin A=\frac{BC}{AC}`

`\sin A=\frac{22}{\sqrt773}`

`A=\sin^{-1}(\frac{22}{\sqrt773})`

`=52.306^\circ`

Similarly, using sine formula in right triangle to find the angle `C` as follows

`\sin C=\frac{AB}{AC}`

`\sin A=\frac{17}{\sqrt773}`

`A=\sin^{-1}(\frac{17}{\sqrt773})`

`=37.694^\circ`

Hence, all three sides & angles of right `\triangle ABC` are

`AB=17, BC=22, AC=\sqrt773,`

`A=52.306^\circ, B=90^\circ, C=37.694^\circ`

Answer 2

Solution: `AB=17,BC=22,AC~~ 27.8` and
`/_A=52.31^0, B=90^0,/_C=37.69^0`

Explanation:

Let Opposite , adjacent sides of the right triangle be

`AB=c=17 and BC=a=22` and `/_B=90^0`

`AC=b` is hypotenuse , In right triangle ,

`a^2+c^2=b^2 or b^2= 17^2+22^2=773` or

`b= sqrt 773 ~~ 27.8:. AC=27.8` unit .

The sine rule states if `a, b and c` are the lengths of the sides

and opposite angles are `A, B and C` in a triangle, then:

`a/sin A = b/sin B=c/sin C :. b/sin B=c/sin C` or

`27.8/sin 90 = 17 /sin C or sin C= 17/27.8 ; [sin 90=1]` or

`sin C~~ 0.611448 or /_C~~ sin^-1( 0.611448)` or

`/_C~~ 37.69^0 :. /_A= 180-(90+37.69)~~ 52.31^0`

Solution: `AB=17,BC=22,AC~~ 27.8` and

`/_A=52.31^0, B=90^0,/_C=37.69^0` [Ans]