SSC CHSL 4 Mar 2018 Shift 3 Paper

Section: English Language

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Question : 25 of 100

Marks: +1, -0

ΔPQR is right angled at Q. If cosecP = 17/15, then what is the value of sinR ?

15/17
8/17
17/8
17/15

Solution:

Given :

\cosecP = 17/15

⇒

\sin P = {QR}/{PR }=15/ 17

Let QR = 15 cm and PR = 17 cm
Thus, in ∆PQR, ⇒

(PQ)^ 2 = (PR) ^2 − (QR) ^2

⇒

(PQ) ^2 = (17)^ 2 − (15) ^2

⇒

(PQ) ^2 = 289 − 225 = 64

⇒

PQ=√64=8

cm
To find : sin R =

{PQ}/ {PR}

=

8/17

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