Question

In an acute angled triangle abc if tan ( A+B-C)=1 and sec ( B+C-A)=2, then find the value of cos ( 4B-3A)=?

Answer 1

Answer:

sqroot 3/2

Step-by-step explanation:

we know tan 45=1

so, A+B-C=45            ......(1)

and sec 60=2

so ,B+C-A=60            .......(2)

(1) +(2)

2B=105

B=105/2

we know A+B+C=180     (as it is a triangle)

A+C=255/2                 (putting value of B)       .......(3)

in (1)

A-C=-15/2           (Putting value of B)         .........(4)

(3) + (4)

2A=120

A=60

putting the values of A and B in cos(4B-3A)

=cos (4*105/2-3*60)

=cos(210-180)

=cos 30

=sqroot3/2