!!!! 100 POINTS !!! How would you go about solving this? Please provide formulae, rules, explanation / working out . Thank you in advance!

The problem involves finding the length of a side (QS) of a triangle using the cosine rule, given two sides and the included angle. The triangle is part of a parallelogram where the area and two sides are given.

Steps:

1. Determining the Angle:

The problem starts by determining the angle SPQ (let's call it angle [tex]x [/tex]) of the parallelogram. The area of a parallelogram is given by the formula [tex] \text{base} \times \text{height} \times \sin(x) [/tex]

By substituting the given values for base (3.8), height (6.1), and area (18 cm²), we can solve for [tex]\sin(x) [/tex]

[tex]

\sin(x) = \frac{18}{3.8 \times 6.1} \approx 0.7765

[/tex]

Since [tex]x [/tex] is acute, we find [tex] x [/tex]using the inverse sine function (arcsin).

[tex]

x = \arcsin(0.7765) \approx 50.94^\circ

[/tex]

2. Using the Cosine Rule:

With angle [tex] x [/tex]known, we proceed to find the length of side QS (let's call it [tex] y [/tex]) using the cosine rule for triangles, which relates the lengths of the sides of a triangle to the cosine of one of its angles.

The cosine rule is expressed as:

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

For our triangle, with sides PS (3.8), PQ (6.1), and QS (which we want to find), and the angle opposite QS as [tex] x [/tex], the cosine rule becomes:

[tex] y^2 = 3.8^2 + 6.1^2 - 2 \times 3.8 \times 6.1 \times \cos(50.94^\circ) [/tex]

By substituting the values, we can solve for [tex] y^2 [/tex]and then find [tex] y [/tex] by taking the square root.

3. Finding "y" :

[tex] y^2 \approx 14.44 + 37.21 - 2 \times 3.8 \times 6.1 \times 0.7765 \approx 51.65 - 28.23 \approx 23.42 [/tex]

[tex]

y \approx \sqrt{23.42} \approx 4.84

[/tex]

When rounding to three significant figures, [tex] y [/tex] is approximately 4.84 cm.

Therefore, the length of QS is approximately 4.84 cm.

msm555 msm555 · Answer 2 · 2024-04-07T05:59:49+02:00

Answer:

QS = 4.74 cm

Step-by-step explanation:

Given:

Parallelogram PQRS with [tex]\sf \angle SPQ [/tex] being acute.
[tex]\sf PQ = 6.1 [/tex] cm
[tex]\sf PS = 3.8 [/tex] cm
Area of the parallelogram: [tex]\sf 18 \textsf{ cm}^2 [/tex]

To find:

Length of [tex]\sf QS [/tex]

Solution:

Calculate the Angle [tex]\sf SPQ [/tex] ([tex]\sf x[/tex]):

Given that the area of the parallelogram is [tex]\sf 18 \textsf{ cm}^2 [/tex], we use the formula for the area of a parallelogram:

[tex]\sf \textsf{Area} = PQ \times PS \times \sin(\textsf{angle between PQ and PS}) [/tex]

[tex]\sf 18 = 3.8 \times 6.1 \times \sin(x) [/tex]

Solve for [tex]\sf \sin(x) [/tex]:

[tex]\sf \sin(x) = \dfrac{18}{3.8 \times 6.1} [/tex]

[tex]\sf \sin(x) = \dfrac{18}{23.18} [/tex]

[tex]\sf \sin(x) \approx 0.7765314926660 [/tex]

Find [tex]\sf x [/tex] (angle [tex]\sf SPQ [/tex]):

we will take inverse of sin.

[tex]\sf x = \sin^(0.7765314926660) [/tex]

[tex]\sf x \approx 50.944090045678^\circ [/tex]

Apply the Cosine Rule to Find [tex]\sf QS [/tex] ([tex]\sf y[/tex]):

Using the cosine rule in triangle [tex]\sf PQS [/tex], where [tex]\sf y = QS [/tex]:

[tex]\sf y^2 = PQ^2 + PS^2 - 2 \times PQ \times PS \times \cos(x) [/tex]

[tex]\sf y^2 = 3.8^2 + 6.1^2 - 2 \times 3.8 \times 6.1 \times \cos(50.944090045678^\circ) [/tex]

[tex]\sf y^2 = 3.8^2 + 6.1^2 - 2 \times 3.8 \times 6.1 \times 0.6428 [/tex]

[tex]\sf y^2 \approx 14.44 + 37.21 - 2 \times 3.8 \times 6.1 \times 0.6300784402737 [/tex]

[tex]\sf y^2 \approx 51.65 - 29.210436491089 [/tex]

[tex]\sf y^2 \approx 22.439563508910 [/tex]

[tex]\sf y \approx \sqrt{22.439563508910} [/tex]

[tex]\sf y \approx 4.7370416410361 [/tex]

[tex]\sf y \approx 4.74 \textsf{ cm(in 3 significant figures)} [/tex]

Therefore, the length of [tex]\sf QS [/tex] is approximately [tex]\sf \boxed{4.74} [/tex] cm.