In ∆ABC cosA+cosC=sinB , then prove that triangle is right angled triangle.
Rishikesh Shinde , 6 Years ago
Grade 11
Trigonometry> In ∆ABC cosA+cosC=sinB , then prove that ...
1 Answers
Arun
Last Activity: 6 Years ago
We have
cosA+cosC = 2cos(A+C)/2cos(A–C)/2 ...i
& sinB = 2sinB/2cosB/2. ...ii
and we know that A+B+C=180°
(A+C)/2=(180°–B)/2=90°–B/2
adding cos on both sides we get
cos(A+C)/2 = cos(90°–B/2) = sinB/2. ...iii
now coming to the question
cosA+cosB = sinB
by equating those formulas
2cos(A+C)/2cos(A–C)/2 = 2sinB/2cosB/2
2sinB/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/2
A=B+C. ...iv
we know that A+B+C = 180°
A+A=180° (by iv )
2A=180°
A=90°
since angle A=90°it is right angled at A
so ABC is a right angled triangle
cosA+cosC = 2cos(A+C)/2cos(A–C)/2 ...i
& sinB = 2sinB/2cosB/2. ...ii
and we know that A+B+C=180°
(A+C)/2=(180°–B)/2=90°–B/2
adding cos on both sides we get
cos(A+C)/2 = cos(90°–B/2) = sinB/2. ...iii
now coming to the question
cosA+cosB = sinB
by equating those formulas
2cos(A+C)/2cos(A–C)/2 = 2sinB/2cosB/2
2sinB/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/2
A=B+C. ...iv
we know that A+B+C = 180°
A+A=180° (by iv )
2A=180°
A=90°
since angle A=90°it is right angled at A
so ABC is a right angled triangle
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