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| Projection Formula |
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| In any triangle ABC, with sides a = BC, b = CA and c = BA, then prove
that a = b cos C + c cos B, b = c cos A + a cos C, c
= a cos B + b cos A |
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| Fig (i) |
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| Fig (ii) |
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| Fig (iii) |
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| Proof: |
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| Let ABC be any triangle and let AD be drawn perpendicular to BC. |
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| (AD = h = altitude of triangle) |
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| In figure (i), we have |
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| BD = x, DC = y |
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| From right-angled triangle ADB, we have |
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| From right-angled triangle ADC, we have |
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| Adding (ii) and (iii), we get |
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| x + y = cos B + b cox C …. (iv) |
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| From (i) and (iv), we have |
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| From figure (ii), we have |
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| BD = x, DC = y |
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| In D ADB, |
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| In D ADC, |
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| From fig (iii), |
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| Hence, a = b cos C + c cos B. |
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| Similarly, other three results can be proved. |
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