If in a triangle ABC right angled at B, AB = 6 units and BC = 8 units

Q) 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 + cos A sin C.

Ans:

Step 1: Let’s make a diagram to better understand the question:

Here Δ ABC is a right angle triangle and ∠B is 90.

we are given AB = 6 units, BC = 8 units

Step 2: By Pythagoras theorem, let’s calculate the value of side AC:

AC² = AB² + BC²

∴ AC² = (6)² + (8)² = 36 + 64

= 100 = (10)²

∴ AC = 10 units

Step 3: From the diagram, let’s calculate the different values:

sin A = $\frac{BC}{AC} = \frac{8}{10} = \frac{4}{5}$

and cos A = $\frac{AB}{AC} = \frac{6}{10} = \frac{3}{5}$

Similarly, sin C = $\frac{AB}{AC} = \frac{6}{10} = \frac{3}{5}$

and cos C = $\frac{BC}{AC} = \frac{8}{10} = \frac{4}{5}$

Step 4: Let’s put the values now:

sin A cos C + cos A sin C

= $\frac{4}{5} \times \frac{4}{5} + \frac{3}{5} \times \frac{3}{5}$

= $\frac{16}{25} + \frac{9}{25} = \frac{16 + 9}{25}$

= $\frac{25}{25}$ = 1

Therefore, the value of sin A cos C + cos A sin C is 1.

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