# The altitude AD of triangle ABC,in which <A is obtuse AD=10 cm. If BD=10cm and CD=10√3 cm, determine <A, AB and AC.

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Since AD is the height of triangle ABC, then the triangles ADB and ADC are right angle triangles.

Since ADB is a right angle triangle and AB is the side opposed to the right angle, hence, AB represents the hypotenuse and you may write the Pythagorean theorem such that:

`AB^2 = AD^2 + DB^2`

`AB^2 = 100 +100 => AB = sqrt(200) => AB = 10sqrt2`

Since ADC is a right angle triangle and AC is the side opposed to the right angle, hence, AC represents the hypotenuse and you may write the Pythagorean theorem such that:

`AC^2 = AD^2 + CD^2`

`AC^2 = 100 + 300 => AC = sqrt(400) => AC = 20`

You should use the law of cosines to find the angle `hat A ` such that:

`BC^2 = AB^2 + AC^2 - 2AB*AC*cos hat A`

`cos hatA = (AB^2 + AC^2 - BC^2)/(2AB*AC)`

`cos hatA = (200 + 400 - (10 + 10sqrt3)^2)/(400sqrt2)`

`cos hatA = (200 + 400 - 100- 200sqrt3 - 300)/(400sqrt2)`

`cos hatA = (200(1- sqrt3))/(400sqrt2)`

`cos hat A = (1 - sqrt3)/2sqrt2 => cos hat A = sqrt2/4 - sqrt6/4`

`cos hat A = cos (pi/3+ pi/4) = cos(pi/3)cos(pi/4) - sin(pi/3)sin(pi/4)`

`cos (pi/3 + pi/4) = (1/2)(sqrt2/2) - (sqrt3/2)(sqrt2/2)`

`cos 7pi/12 = sqrt2/4 - sqrt6/4`

**Hence, evaluating the lengths of the sides AB and AC yields `AB = 10sqrt2` and `AC = 20` and the angle `hat A = 7pi/12` .**

PLZZZZZZZ.....

is there a simple way?????????