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Hemanth Kumar

· started a discussion

· 1 Months ago

the diagram is wrong (Question: 257141.0 )

SM is perpendicular to PQ ..... SN is perpendicular to PR.... the answer should be = 2/ (square root 3)

Question:

An isosceles triangle PQR is right angled at Q. S is a point inside the triangle PQR, M and N are the foot of the perpendiculars drawn from S on the sides PQ and PR respectively. If PM = 8 cm, PN = 6 cm and ∠QPS = 15°, sin75° = ?

Options:
A)

223    

B)

43    

C)

\(\frac{3√3}{8}\) 

D)

833    

Solution:

Ans: (c)


In ∆ PMS

Sin 75° = \(\cfrac{PS}{8}\)

And in ∆ PNS

Sin 60° = \(\cfrac{PS}{6}\)

PS = 6 × \(\cfrac{√3}{2}\) = 3√3 cm

∴ sin75° = \(\cfrac{PS}{8}\) = \(\cfrac{3√3}{8}\)

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