Question

Cos A= Sin B- sin C then prove that it is a right angled triangle

Answer 1

Answer:

Step-by-step explanation:

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

As per 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