Question

the altitude AD of a ABC in which angle A is obtuse is 10 cm.
If BD = 10 cm and CD = 10√3 cm, determine angle A and AB and AC ​

Answer 1

Step-by-step explanation:

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 .