SOLVE PLZ AS SOON AS POSSIBLE I BEG U1.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.

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:

[tex]\tan60^o= \frac{AC}{CL} \\ \\ \Rightarrow AC=CL\tan60^o \ . \ . \ . \ (1)[/tex]
and
[tex]\tan30^o= \frac{AC}{18+CL} \\ \\ \Rightarrow AC=(18+CL)\tan30^o \ . \ . \ . \ (2)[/tex]

From (1) and (2):

[tex]CL\tan60^o=(18+CL)\tan30^o \\ \\ \Rightarrow \sqrt{3} CL=18\left( \frac{1}{ \sqrt{3} } \right)+\left( \frac{1}{ \sqrt{3} } \right)CL \\ \\ \Rightarrow \left( \sqrt{3} - \frac{1}{ \sqrt{3} } \right)CL=18\left( \frac{1}{ \sqrt{3} } \right) \\ \\ \Rightarrow \frac{2}{ \sqrt{3} } CL=18\left( \frac{1}{ \sqrt{3} } \right) \\ \\ \Rightarrow CL= \frac{18}{2} =9m[/tex]



Part B:

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

Then:

[tex]\tan30^o= \frac{6}{CD} \\ \\ \Rightarrow CD= \frac{6}{\tan30^o} \\ \\ = \frac{6}{\left( \frac{1}{ \sqrt{3} } \right)} =6 \sqrt{3} [/tex]

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

Thus:

[tex]\tan60^o= \frac{BD}{CD} = \frac{BD}{6 \sqrt{3} } \\ \\ \Rightarrow BD=6 \sqrt{3}\tan60^o \\ \\ =6 \sqrt{3}( \sqrt{3} )=6(3)=18cm[/tex]



Part C:


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

Thus:

[tex]\cos75^o= \frac{AC}{AB} = \frac{AC}{12} \\ \\ AC=12\cos75^o[/tex]

The area of △ABC is given by:

[tex] \frac{1}{2} \times12\times AC\times\sin75^o=6(12\cos75^o)(\sin75^o) \\ \\ =72 \frac{\left(\sin2(75^o)\right)}{2} =36\sin150^o=36\sin30^o=36(0.5) \\ \\ =\bold{18cm^2}[/tex]