) In the right △ABC with m∠C=90°, m∠B=75°, and AB = 12 cm. Find the area of △ABC.
(no sin and cos pleasee)
(show work too)

9514 1404 393

  18 cm^2

Step-by-step explanation:

Under the given constraints, the way to find the area of this triangle is to use a triangle solver app or web site. The attached shows the area is 18 cm^2.

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A simple way to find the area is to use the cosine function to find side AC adjacent to the 75° angle. Then use the formula for area that uses the sine function.

  AC = AB·cos(75°)

  Area = (1/2)AB·AC·sin(75°)

  Area = (1/2)(AB²)cos(75)sin(75°)

  = (1/4)AB²·sin(150°) . . . using the double-angle identity

  = (1/4)(12 cm)²(1/2)

  Area = 18 cm²

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The useful relations are ...

  Cos = Adjacent/Hypotenuse

  cos(75°) = AC/AB   ⇒   AC = AB·cos(75°)

and ...

  Area = (1/2)pq·sin(α) . . . . for two sides p and q with angle α between them

  Area = (1/2)AB·AC·sin(75°)

Another identity we used here is ...

  sin(2α) = 2sin(α)cos(α)   ⇔   sin(α)cos(α) = (1/2)sin(2α)

Putting these together, the formula for area we used was effectively ...

  A = (1/4)AB²sin(2·75°)

tube or goog would be a great resource to with that question