Question

The altitude AD of triangle ABC , in which angle A obtuse and AD = 10 cm . if BD = 10 cm and CD = 10 square root 3 cm , determine the angle A

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°

Answer 2

The altitude AD of triangle ABC , in which angle A obtuse and AD = 10 cm . if BD = 10 cm and CD = 10 square root 3 cm , determine the angle A

Answer:

determine the angle A

Angle A is 105°

Step-by-step explanation:

Given statement

Angle A is an obtuse and AD⊥BC So,∠ADC=90°

Thus, in ΔADC

tan ∠ACD  = AD/CD=10/10/√3=1/3√=tan 30°

⇒∠ACD = 30°

and, in ΔADB

tan ∠ABD=AD/BD=10/10=1=tan 45°

⇒∠ABD=45°

Now,

∠A=∠ABD+∠ACD

⇒∠A=45°+60°=105°

Angle A is 105°