Question

If in a triangle ABC right angled at B, AB = 6 units and BC = 8 units, then find
the value of sin A . cos C.​

Answer 1

Answer:

0.64.........................

Answer 2

Answer:

Solution:

Given In ∆ABC, <ABC = 90°

AB = 6 units , BC = 8 units

By Phythogarian theorem:

i) AC² = AB²+BC²

= 6²+8²

= 36+64

= 100

=> AC = √100

=> AC = 10 units.

ii)sinA = BC/AC = 8/10

cosC = BC/AC = 8/10

cosA = AB/AC = 6/10

sinC = AB/AC = 6/10

Now ,

sinAcosC+cosAsinC

= \frac{8}{10}\times\frac{8}{10}+\frac{6}{10}\times\frac{6}{10}

10

8

×

10

8

+

10

6

×

10

6

= \frac{64}{100}+\frac{36}{100}

100

64

+

100

36

= \frac{64+36}{100}

100

64+36

= \frac{100}{100}

100

100

= 11

Therefore,

sinAcosC+cosAsinC =1

hope it helps you