Question

In triangle PQR, PR=7cm, RQ=9cm and angle PQR=108°, calculate the length of PQ. Pls explain in a simple way

Answer 1

To find the length of side PQ in triangle PQR, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides \( a \), \( b \), and \( c \), and an angle \( C \) opposite side \( c \):

\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

In triangle PQR:
- Let \( a = PR = 7 \, \text{cm} \)
- Let \( b = RQ = 9 \, \text{cm} \)
- Let \( C = \angle PQR = 108^\circ \)
- Let \( c = PQ \) (which we want to find)

Using the Law of Cosines:

\[
PQ^2 = PR^2 + RQ^2 - 2 \cdot PR \cdot RQ \cdot \cos(\angle PQR)
\]

Substituting the values:

\[
PQ^2 = 7^2 + 9^2 - 2 \cdot 7 \cdot 9 \cdot \cos(108^\circ)
\]

Calculating each term:

1. \( 7^2 = 49 \)
2. \( 9^2 = 81 \)
3. \( PR \cdot RQ = 7 \cdot 9 = 63 \)

Now we need to find \( \cos(108^\circ) \). From trigonometric tables or a calculator, we know:

\[
\cos(108^\circ) \approx -0.309
\]

Now plug in this value:

\[
PQ^2 = 49 + 81 - 2 \cdot 63 \cdot (-0.309)
\]

Calculating:

1. \( 49 + 81 = 130 \)
2. \( -2 \cdot 63 \cdot (-0.309) = 2 \cdot 63 \cdot 0.309 \approx 39.078 \)

Now, add these results:

\[
PQ^2 \approx 130 + 39.078 = 169.078
\]

To find PQ, take the square root:

\[
PQ \approx \sqrt{169.078} \approx 13.0 \, \text{cm}
\]

Thus, the length of side PQ is approximately **13.0 cm**.