Question

The altitude AD of triangle ABC in which angle A is acute is 10 cm. if BD = 10 cm and CD = 10 √3 , what is angle of BAC?​

Answer 1

From Pythagoras theorem,

AB = √(AD² + BD²) = 10√2 cm

AC = √(AD² + CD²) = 20cm

use cosine theorem,

cos angle BAD = (AB² + AD² - BD²)/2AB.AD

= (200 + 100 - 100)/2(10√2)(10)

= 200/200√2 = 1/√2

so, cos angle BAD = 1/√2 = cos45°

angle BAD = 45°

again, cos angle DAC = (AD² + AC² - CD²)/2AD.AC

= (100 + 400 - 300)/2(10)(20)

= 200/400 = 1/2 = cos60°

hence, angle CAD = 60°

so, angle BAC = 45° + 60° = 105°