Question

If $\sin A = -\frac{24}{25}$, $\cos B = \frac{15}{17}$, A does not belong to 4\textsuperscript{th} quadrant and B does not belong to 1\textsuperscript{st} quadrant then $(A + B)$ lies in the quadrant

• A. 1\textsuperscript{st} quadrant
• B. 2\textsuperscript{nd} quadrant
• C. 3\textsuperscript{rd} quadrant
• D. 4\textsuperscript{th} quadrant

Hint:

To determine the quadrant of a sum of angles like $(A+B)$, find the signs of $\sin(A+B)$ and $\cos(A+B)$. Remember the quadrant sign rules: Q1(+,+), Q2(-,+), Q3(-,-), Q4(+,-) for $(\cos, \sin)$.

Answer 1

The Correct Option is C

Step 1: Determine the quadrant of angle A.
We are given $\sin A = -24/25$, which is negative. The sine function is negative in the 3rd and 4th quadrants.
The problem states that A does not belong to the 4th quadrant. Therefore, A must be in the 3rd quadrant.
In the 3rd quadrant, cosine is also negative. $\cos A = -\sqrt{1-\sin^2 A} = -\sqrt{1 - (-\frac{24}{25})^2} = -\sqrt{1 - \frac{576}{625}} = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.
Step 2: Determine the quadrant of angle B.
We are given $\cos B = 15/17$, which is positive. The cosine function is positive in the 1st and 4th quadrants.
The problem states that B does not belong to the 1st quadrant. Therefore, B must be in the 4th quadrant.
In the 4th quadrant, sine is negative. $\sin B = -\sqrt{1-\cos^2 B} = -\sqrt{1 - (\frac{15}{17})^2} = -\sqrt{1 - \frac{225}{289}} = -\sqrt{\frac{64}{289}} = -\frac{8}{17}$.
Step 3: Determine the signs of $\cos(A+B)$ and $\sin(A+B)$ to find the quadrant of $A+B$.
$\cos(A+B) = \cos A \cos B - \sin A \sin B = (-\frac{7}{25})(\frac{15}{17}) - (-\frac{24}{25})(-\frac{8}{17}) = -\frac{105}{425} - \frac{192}{425} = -\frac{297}{425}$. (Negative)
$\sin(A+B) = \sin A \cos B + \cos A \sin B = (-\frac{24}{25})(\frac{15}{17}) + (-\frac{7}{25})(-\frac{8}{17}) = -\frac{360}{425} + \frac{56}{425} = -\frac{304}{425}$. (Negative)
Since both $\sin(A+B)$ and $\cos(A+B)$ are negative, the angle $(A+B)$ lies in the 3rd quadrant.