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SSC CHSL 7 Mar 2018 Shift 2 Paper

Section: English Language

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Question : 25 of 100

Marks: +1, -0

ΔDEF is right angled at E. If cotD = 5/12, then what is the value of sinF ?

Your Answer:

Solution:

Given :

\cot D = \frac{5}{12}

Also,

\cot D = \frac{DE}{EF} = \frac{5}{12}

Let DE = 5 cm and EF = 12 cm

Thus, in ∆DEF , ⇒

(DF)^2 = (DE)^2 + (EF)^2

⇒

(DF)^2 = (5)^2 + (12)^2

⇒

(DF)^2 = 25 + 144 = 169

⇒

DF = \sqrt{169} = 13

cm

To find :

\sin F = \frac{DE}{DF}

=

\frac{5}{13}

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