Question

Please answer this math question asap!! 30 points and brainliest!!
BRAINLIEST ONLY IF YOU SHOW YOUR WORK!
MUST SHOW WORK!

Answer 1

From image we Can see That :-

→ EF = Smaller Building = 80ft

→ CD = Distance of Taller building From Smaller Building .

→ Angle ACD = 49°

→ Angle DCB = 62°

→ DB = EF = 80ft.

Solution :-

in ∆DCB, we Have :-

→ Angle DCB = 62°

→ DB = EF = 80ft.

So,

→ Tan@ = Perpendicular / Base

→ Tan62° = DB/CD

→ CD = (80/Tan62°)

→ CD = (80/1.88)

→ CD = 42.536 ft.

___________________________

Now ,

in Rt.∆ADC we Have :-

→ CD = 42.536ft.

→ Angle ACD = 49°

So,

Again, Tan@ = Perpendicular / Base

→ Tan49° = AD/CD

→ AD = Tan49° * CD

→ AD = Tan49° * 42.536

→ AD = 1.15 * 42.536

→ AD = 48.916ft.

_____________________

So,,

==>> Height of Taller Building = AD + DB = 48.916 + 80 = 128.916 ft ≈ 129ft .

Hence, Height of Taller Building is 129ft.

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

★ Given :

Height of building (EA or BC) = 80 ft

Angle of elevation (1) = 49°

Angle of elevation (2) = 62°

__________________________

★ To Find :

We have to find the height of taller building (BD)

__________________________

★ Solution :

In ∆ BCE

$\sf{→tan \: 62^{\circ} = \frac{BC}{EC}} \\ \\ \sf{→EC = \frac{80}{tan \: 62^{\circ}}.....(1)}$

$\rule{150}{2}$

Now, In ∆ DEC

$\sf{→tan \: 49^{\circ} = \frac{DC}{EC}} \\ \\ \sf{→EC = \frac{DC}{tan \: 49^{\circ}} .......(2)}$

$\rule{200}{2}$

From equation (1) and (2).

$\sf{→\frac{80}{tan \: 62^{\circ}} = \frac{DC}{tan \: 49^{\circ}}} \\ \\ \sf{→DC = \frac{80}{tan \: 62^{\circ}} \times tan \: 49^{\circ}} \\ \\ \sf{→DC = \frac{80 \times 1.5037}{1.88073}} \\ \\ \sf{→DC = 63.96 \: ft}$

$\rule{200}{1}$

Now,

Height of tower = BC + DC

→ BD = 80 + 63.96

→ BD = 143.96 ft

$\large{\star{\boxed{\sf{Height \: of \: tower = 143.96 \: ft}}}}$