In $Δ$ PQR, right angled at Q, PQ=4 cm and RQ=3 cm. Find the values of sin P, sin R, sec P and sec R.

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Solution

ΔPQR, right angled at Q.

Let x be the hypotenuse
By applying Pythagoras

$P R^{2} = P Q^{2} + Q R^{2} x^{2} = 4^{2} + 3^{2} x^{2} = 16 + 9 ∴ x^{2} = \sqrt{25} = 5$

Now we need to find $s i n P$ , $s e c P$ , $s e c R$

At LP, opposite side = 3cm
Adjacent side = 4cm
Hypotenuse = 5cm
$s i n P = \frac{o p p o s i t e s i d e}{H y p o t e n u s e} = \frac{3}{5}$
$s e c P = \frac{H y p o t e n u s e}{A d j a c e n t s i d e} = \frac{5}{4}$

At LK, opposite side = 4cm
Adjacent side = 3cm
Hypotenuse = 5cm
$s i n R = \frac{4}{5}$
$s e c R = \frac{5}{3}$

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