# Thread: Trig Math Help

1. ## Trig Math Help

How would someone begin to figure a question like this out? If anyone can help me solve this problem it would be greatly appreciated.

2. Originally Posted by gg_gone
How would someone begin to figure a question like this out? If anyone can help me solve this problem it would be greatly appreciated.
$\sin{30} = \frac{30}{BC}$

$\tan{30} = \frac{30}{BA}$

$\sin{52} = \frac{30}{BD}$

$\tan{52} = \frac{30}{AD}$

3. angle BDA = 180 - 128 = 52 degrees

tan BCD = tan 30 = 30/AC = 30/(AD + CD) ---- (1)

tan BDA = tan 52 = 30/AD ---- (2)

from (1)

(AD + CD) = 30/tan 30,

AD = 30/tan 30 - CD ---- (3)

from (2)

AD = 30/ tan52 ---- (4)

equate equation (3) and (4)

30/tan 30 - CD = 30/ tan 52,

solving for CD, we have

CD = 30/tan 30 - 30/ tan 52

CD = 28.52 m

then AD = 30/ tan52 = 23.44 m

A) Area of ABC,

AC = AD + CD

AC = 23.44 + 28.52 = 51.96 m

BC = [(30)^2 + (51.96)^2)]^(1/2) = 60 m

Area = (1/2)(AB)(AC)

= (1/2)(30)(51.96) = 779.4 m^2

Perimeter = AB + AC + BC

= 30 + 51.96 + 60 = 141.96 m

You solve the rest, ok

4. gg_gone, where are you?