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In triangle ABC, AC =24, the measure of angle A = 30 degree , and the measure of angle B = 45 degree . Find the area of this triangle. (Hint: Draw a picture and include the altitude from vertex C to the side AB.)

Answer :

Answer:

142.932

Step-by-step explanation:

given that in triangle ABC, AC =24

Angle A = 30 and Angle B =45

If we draw altitude from C, we get

h = 24sin 30 = 12

Since BC = h (45, 45,90 triangle)

BC=12

By sine formula for triangle

[tex]\frac{12}{sin30} =\frac{AB}{sin(180-45-30)} \\AB = 24 sin 105 = 23.1822[/tex]

Area of triangle

= [tex]1/2 (23.1822)(12)\\= 142.932[/tex]

The  area of this triangle is 142.932.

  • The calculation is as follows:

In the case when  we draw altitude from C, we get

So,

h = 24sin 30 = 12

Since BC = h (45, 45,90 triangle)

BC=12

By using sine formula for triangle

[tex]\frac{12}{sin30} = \frac{AB}{sin(180-45-30)}[/tex]

AB = 24sin105

= 23.1822

Now

= 0.5(23.1822) ( 12)

= 142.9232

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