Question

. In triangle ABC, right angled at B, if tan A = 6/8, then find the value of sin A. cos C + cos A. sin C

Answer 1

Answer:

cosA cos C - sinA sin C= cos(a+c)= cos (30+60)=0. diavinad8 and 20 ... In right angled triangle ABC,. tanA= 1/root3 ... And by angle sum property in triangle ABC , we get C=60 degrees.

Answer 2

Answer:

sinA=3/5, cosC+cosA=9/5, sinC=4/5

Step-by-step explanation:

tanA=6/8

tanA also=P/B

Therefore,

P=6

and B=8

By phythagoras theorem,

H²=B²+P²

H²=(8)²+(6)²

H²=64+36

H²=100

H=√100

H=10

sinA=P/H

sinA=6/10

sinA=3/5

cosC=B/H

cosC=6/10

cosC=3/5

cosA=8/10

cosA=4/5

therefore,

cosC+cosA=3/5+4/5

cosC+cosA=3+4/5

cosC+cosA=9/5

sinC=P/H

sinC=8/10

sinC=4/5