SSC-IMD-SA-QUANT-MOCK-01 – Question 1

Question:

If \(\sin P = \dfrac{5}{13}\) and \(\cos Q = \dfrac{12}{13}\), find \(\sin P \cos Q + \cos P \sin Q\):

यदि \(\sin P = \dfrac{5}{13}\) और \(\cos Q = \dfrac{12}{13}\) है, तो \(\sin P \cos Q + \cos P \sin Q\) का मान ज्ञात कीजिए:

Options:

  • 60/169
  • 13/13
  • 120/169
  • 12/13
  • 60/169
  • 13/13
  • 120/169
  • 12/13

Correct Answer: c)

Explanation:

From \(\sin P = \dfrac{5}{13}\), \(\cos P = \sqrt{1 – (5/13)^2} = \dfrac{12}{13}\).
From \(\cos Q = \dfrac{12}{13}\), \(\sin Q = \sqrt{1 – (12/13)^2} = \dfrac{5}{13}\).
Then \(\sin P \cos Q + \cos P \sin Q = \left(\dfrac{5}{13}\right)\left(\dfrac{12}{13}\right) + \left(\dfrac{12}{13}\right)\left(\dfrac{5}{13}\right) = \dfrac{60}{169} + \dfrac{60}{169} = \dfrac{120}{169}\).

\(\cos P = \dfrac{12}{13}\) तथा \(\sin Q = \dfrac{5}{13}\).
\(\sin P \cos Q + \cos P \sin Q = \left(\dfrac{5}{13} \times \dfrac{12}{13}\right) + \left(\dfrac{12}{13} \times \dfrac{5}{13}\right) = \dfrac{60}{169} + \dfrac{60}{169} = \dfrac{120}{169}\)।

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