Q. 155.0( 1 Vote )


In the figure be

Answer :


https://gs-post-images.grdp.co/user_files/63094/15047/images/extra-9_files/Image1110.gif

Solution:

In right angled ΔPQR,

P = 90°

PQR = 20°

sec θ = hypotenuse / adjacent


sec 20° =
QR / PQ


1.0642 =
QR / 3


QR = 1.0642 × 3

= 3.1926


QR = ST = 3.2 units


In right angled Δ ACD,

CAD = 42° 6'


cot 42° 6' =
AD / DC


1.1067 =
AD / 1.5

AD = 1.5 × 1.1067

= 1.66 units

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