Question

If cosA=sinB-cosC is in the triangle, then show that the triangle is right angled. ​

Answer 1

Answer:

Step-by-step explanation:

Given, cos(A) = sin(B) - cos(C) ► cos(A) + cos(C) = sin(B)

LHS = 2cos[(A+C)/2]cos[(A–C)/2]

RHS = sinB = 2sin(B/2)cos(B/2)

Since A+B+C=180°, (A+C)/2 = (180°–B)/2 = 90°– B/2

or cos[(A+C)/2] = cos(90°–B/2) = sin(B/2)

Hence, 2sin(B/2)cos[(A–C)/2] = 2sin(B/2)cos(B/2)

or cos[(A–C)/2] = cos(B/2) since B ≠ 0

Hence, (A–C)/2 = B/2 or A = B+C or 2A =180° or the Tr. ABC is right at A.

Answer 2

Step-by-step explanation:

We have,

$\rm \cos A = \sin B -\cos C$

$\rm \cos A + \cos C = \sin B$

$\rm \implies 2cos \bigg( \dfrac{A+C}{2} \bigg) cos \bigg( \dfrac{A-C}{2} \bigg) = \sin \dfrac{B}{2} \cos \dfrac{B}{2}$

$\rm \implies \dfrac{A+B+C}{2}= \dfrac{π}{2}$

$\rm \implies \dfrac{A+C}{2}= \dfrac{π}{2}=\dfrac{B}{2}$

$\rm \implies \cos \bigg( \dfrac{A+C}{2} \bigg) = \sin \dfrac{A}{2}$

$\rm \implies 2 \sin \dfrac{B}{2}\cos \bigg( \dfrac{A-C}{2} \bigg) = 2 \sin \dfrac{B}{2} \cos \dfrac{B}{2}$

$\rm \implies \dfrac{A-C}{2}= \dfrac{B}{2}$

$\rm \implies A-C=B$

$\rm \implies A=B+C$

$\rm \implies A+B+C= π$

$\rm \implies A+A= π \Rightarrow 2A = π \Rightarrow A = \dfrac{π}{2}$