Question

In the given triangle ABC which is right-angled at B, if sin A =$\frac{12}{13}$, then find the value of (sinA cosA + sinC cosC).

• A. (120/169)
• B. (60/169)
• C. (12/13)
• D. None of the options

Answer 1

The correct option is A

(120/169)

Here,Sin A = $\frac{BC}{AC}$

Let $\frac{BC}{AB}$ = k , where k is a positive number, such that BC=12k and AC=13k

⇒ $A{C}^2=B{C}^2+A{B}^2$

⇒$(13k{)}^2=(12k{)}^2+(AB{)}^2$

⇒$A{B}^2=169k^2-144k^2=25k^2$

⇒AB = 5k

So, now we can get the required trigonometric ratios:

⇒ cos A = $\frac{AB}{AC}=\frac{5k}{13k}=\frac{5}{13}$

⇒ sin C = $\frac{AB}{AC}=\frac{5k}{13k}=\frac{5}{13}$

⇒ cos c = $\frac{BC}{AC}=\frac{12k}{13k}=\frac{12}{13}$

So, sinA cosA + sin C cos C = $\frac{12}{13}\times \frac{5}{13}+\frac{5}{13}\times \frac{12}{13}$

⇒ sinA cosA + sin C cos C = $\frac{60}{169}+\frac{60}{169}=\frac{120}{169}$.