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

Answer 1

Answer:

∠A = 105°

Step-by-step explanation:

Given, BD = AD = 10 cm, CD = 10√3.

(i) ΔBAD:

tan ∠ABD = BD/AD

                = 10/10

                = 1

tan ∠BAD = 1

∠BAD = 45°

(ii) ΔCAD:

tan ∠CAD = CD/AD

                 = 10√3/10

                 = √3

tan ∠CAD = √3

∠CAD = 60°

From (i) & (ii), we have

∠BAC = ∠BAD + ∠CAD

          = 45° + 60°

          = 105°

Therefore, ∠A = 105°

Hope it helps!

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

__________________________________________________

Δ ABC is a right triangle.

Right angle at D

AD = 10cm

BD = 10 cm

CD = 10√3 cm

∴ Tan ∠BAD = BD/AD

tan∠BAD = 10/10 ⇒ 1

tan∠BAD = tan 45°

∠BAD = 45°.  ---- ( 1 )

ΔACD is a right triangle.

Right angled at D, such that :-

AD = 10cm

CD = 10√3 cm

∴ tan ∠CAD = CD/AD

tan ∠CAD = 10√3/10 ⇒ √3

tan ∠CAD = tan 60°

∠CAD = 60°   ---- ( 2 )

FROM (1) AND (2) WE HAVE :-

∠BAC = ∠BAD + ∠CAD

⇒ 45° + 60°

⇒ 105°

__________________________________________________