Question

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

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

In ΔPQR

∠Q=90°

Perpendicular = PQ = 4cm

Base = QR = 3cm

To find hypotenuse i.e. PR we will use Pythagoras theorem :

$Hypotenuse^{2} =Perpendicular^{2} +Base^{2}$

$PR^{2} =PQ^{2} +QR^{2}$

$PR^{2} =4^{2} +3^{2}$

$PR^{2} =16+9$

$PR^{2} =25$

$PR =\sqrt{25}$

$PR =5$

Now Hypotenuse = 5 cm

Now Using trigonometric ratio :

$sin\theta = \frac{Perpendicular}{Hypotenuse}$

Since $\theta = \angle P$

So perpendicular = QR

So, $sinP = \frac{QR}{PR}$

$SinP = \frac{3}{5}$

Now For Sin R

Perpendicular = PQ

$sinR = \frac{PQ}{PR}$

$SinR= \frac{4}{5}$

$Sec\theta = \frac{Hypotenuse}{Base}$

$Sec R = \frac{Hypotenuse}{Base}$

For Sec R base is QR

So, $Sec R = \frac{PR}{QR}$

$Sec R = \frac{5}{3}$