In $△ P Q R$ , right angled at $Q$ . If $P Q = 5 cm$ and $R Q = 10 cm$ find ${sin}^{2} P$ .

Open in App

Solution

In $△ P Q R,$ we will find the length of the side $P R$ by using Pythagoras theorem,
$\begin{matrix} P R^{2} & = P Q^{2} + Q R^{2} = 25 + 100 = 125 P R & = \sqrt{125} = 5 \sqrt{5} cm \end{matrix}$

Now,
${sin}^{2} P = \frac{{Q R}^{2}}{{P R}^{2}}$
$= \frac{100}{125}$
$= \frac{4}{5}$

Suggest Corrections

Similar questions

In $△$ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find the value of sin P, cos P and tan P.

Question 16
In the given figure, PQ $⊥$ RQ, PQ = 5 cm and QR = 5 cm. Then, $△ P Q R$ is

a) a right-angled triangle but not isosceles
b) an isosceles right-angled triangle
c) isosceles but not a right-angled triangle
d) neither isosceles nor right-angled triangle

PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

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.

Join BYJU'S Learning Program

In △PQR, right angled at Q. If PQ=5 cm and RQ=10 cm find sin2P.

In △PQR, we will find the length of the side PR by using Pythagoras theorem,PR2=PQ2+QR2=25+100=125PR=√125=5√5 cmNow,sin2P=QR2PR2 =100125 =45

In $△ P Q R$ , right angled at $Q$ . If $P Q = 5 cm$ and $R Q = 10 cm$ find ${sin}^{2} P$ .

In $△ P Q R,$ we will find the length of the side $P R$ by using Pythagoras theorem,
$\begin{matrix} P R^{2} & = P Q^{2} + Q R^{2} = 25 + 100 = 125 P R & = \sqrt{125} = 5 \sqrt{5} cm \end{matrix}$

Now,
${sin}^{2} P = \frac{{Q R}^{2}}{{P R}^{2}}$
$= \frac{100}{125}$
$= \frac{4}{5}$