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`        In ∆ABC cosA+cosC=sinB , then prove that triangle is right angled triangle.`
one year ago

```							We have cosA+cosC = 2cos(A+C)/2cos(A–C)/2 ...i& sinB = 2sinB/2cosB/2. ...iiand we know that A+B+C=180°(A+C)/2=(180°–B)/2=90°–B/2adding cos on both sides we get cos(A+C)/2 = cos(90°–B/2) = sinB/2. ...iiinow coming to the question cosA+cosB = sinBby equating those formulas 2cos(A+C)/2cos(A–C)/2 = 2sinB/2cosB/22sinB/2cos(A–C)/2 = 2sinB/2cosB/2 (by III) cancelling 2sinB/2 on both sides cos(A–C)/2 = cosB/2(A–C)/2 = B/2A=B+C. ...ivwe know that A+B+C = 180° A+A=180° (by iv ) 2A=180°A=90°since angle A=90°it is right angled at Aso ABC is a right angled triangle
```
one year ago
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