Question

SOLVE PLZ AS SOON AS POSSIBLE I BEG U

1.Given: ∆ABC, m∠C = 90°
m∠ABC = 30°

AL
- ∠ bisector, LB = 18m
Find: CL

2.Given: ΔАВС, m∠ACB = 90°

CD
⊥
AB
, m∠ACD = 30°
AD = 6 cm.
Find: BD
3.
In the right △ABC with m∠C = 90°, m∠A = 75°, and AB = 12 cm. Find the area of △ABC.

Answer 1

Part A:

Given : ∆ABC, m∠C = 90° m∠ABC = 30°

This meanc that m∠BAC = 60°

Given that AL is an ∠ bisector, this means that m∠CAL = 30° and m∠CLA = 60°

Given that LB = 18m, then:


`\tan60^o= (AC)/(CL) \\ \\ \Rightarrow AC=CL\tan60^o \ . \ . \ . \ (1)`
and

`\tan30^o= (AC)/(18+CL) \\ \\ \Rightarrow AC=(18+CL)\tan30^o \ . \ . \ . \ (2)`

From (1) and (2):


`CL\tan60^o=(18+CL)\tan30^o \\ \\ \Rightarrow √(3) CL=18\left( (1)/( √(3) ) \right)+\left( (1)/( √(3) ) \right)CL \\ \\ \Rightarrow \left( √(3) - (1)/( √(3) ) \right)CL=18\left( (1)/( √(3) ) \right) \\ \\ \Rightarrow (2)/( √(3) ) CL=18\left( (1)/( √(3) ) \right) \\ \\ \Rightarrow CL= (18)/(2) =9m`



Part B:

Given: ΔАВС, m∠ACB = 90°; CD⊥AB; m∠ACD = 30° and AD = 6 cm.

Then:


`\tan30^o= (6)/(CD) \\ \\ \Rightarrow CD= (6)/(\tan30^o) \\ \\ = (6)/(\left( (1)/( √(3) ) \right)) =6 √(3)`

Given that m∠ACB = 90° and m∠ACD = 30°, then m∠BCD = 60°

Thus:


`\tan60^o= (BD)/(CD) = (BD)/(6 √(3) ) \\ \\ \Rightarrow BD=6 √(3)\tan60^o \\ \\ =6 √(3)( √(3) )=6(3)=18cm`



Part C:


Given △ABC with m∠C = 90°, m∠A = 75°, and AB = 12 cm.

Thus:


`\cos75^o= (AC)/(AB) = (AC)/(12) \\ \\ AC=12\cos75^o`

The area of △ABC is given by:


`(1)/(2) *12* AC*\sin75^o=6(12\cos75^o)(\sin75^o) \\ \\ =72 (\left(\sin2(75^o)\right))/(2) =36\sin150^o=36\sin30^o=36(0.5) \\ \\ =\bold{18cm^2}`