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never [62]
3 years ago
6

CosA + cosB + cosC = 1 + 4sinA/2 sinB/2 sinC/2

Mathematics
1 answer:
bulgar [2K]3 years ago
6 0

\cos(A)  +  \cos( B )  +  \cos(C)  \\  = 2 \cos( \frac{A + B}{2} )  \cos( \frac{ A- B}{2} )  + 1 - 2 { \sin }^{2}  \frac{C}{2}  \\  = 1 + 2\cos( \frac{\pi - C}{2} )   \cos( \frac{A - B}{2} )  - 2 { \sin }^{2}  \frac{C}{2}  \\  = 1 + 2 \sin \frac{C}{2}  \cos( \frac{A - B}{2} )  - 2 { \sin }^{2}  \frac{C}{2}  \\  = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \sin \frac{C}{2}]  \\  =  1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \cos( \frac{A  +  B}{2} )] \\  = 1 + 2 \sin\frac{C}{2} \times 2 \sin( \frac{ \frac{A + B}{2}  +\frac{A  -  B}{2} }{2} )  \sin( \frac{ \frac{A + B}{2}   - \frac{A  -  B}{2} }{2} )  \\  = 1 + 4 \sin \frac{A}{2}  \sin \frac{B}{2}  \sin \frac{C}{2}

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using the pythagorean theorem we have that

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Then, the area of the triangle is 6*h/2 = 3h = 15.59 cm^2.

Now we calculate the area of the other 3 triangles, notice that those triangles have the same base and height so we will calculate for one of them and multiply by 3. From the image we know that the height is 15cm and the base is 6 cm so the area is 45cm^2, and 45*3 cm^2 = 135cm^2.

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