Question

Please show the working and answer. you can take a picture for the working.

Answer 1

(a) The area of the triangle is approximately 39.0223 cm²

(b) ∠SQR is approximately 55.582°

Explanation:

(a) By sin rule, we have;

SQ/(sin(∠SPQ)) = PQ/(sin(∠PSQ)), which gives;

5.4/(sin(52°)) = 6.8/(sin(∠PSQ))

∴ (sin(∠PSQ)) = (6.8/5.4) × (sin(52°)) ≈ 0.9923

∠PSQ = sin⁻¹(0.9923) ≈ 82.88976°

Similarly, we have;

5.4/(sin(52°)) = SP/(sin(180 - 52 - 82.88976))

Where, 180 - 52 - 82.88976 = ∠PQS = 45.11024

SP = 5.4/(sin(52°))×(sin(180 - 52 - 82.88976)) ≈ 4.8549

Given that RS : SP = 2 : 1, we have;

RS = 2 × SP = 2 × 4.8549 ≈ 9.7098

We have by cosine rule,
`\overline {RQ}`² =
`\overline {SQ}`² +
`\overline {SR}`² - 2 ×
`\overline {SQ}` ×
`\overline {SR}` × cos(∠QSR)

∠QSR and ∠PSQ are supplementary angles, therefore;

∠QSR = 180° - ∠PSQ = 180° - 82.88976° = 97.11024°

∠QSR = 97.11024°

Therefore;

`\overline {RQ}`² = 5.4² + 9.7098² - 2 × 5.4×9.7098× cos(97.11024)

`\overline {RQ}`² ≈ 136.42

`\overline {RQ}` = √(136.42) ≈ 11.6799

The area of the triangle = 1/2 ×
`\overline {PQ}` ×
`\overline {PR}` × sin(∠SPQ)

By substituting the values, we have;

1/2 ×
`\overline {PQ}` ×
`\overline {PR}` × sin(∠SPQ)

1/2 × 6.8 × (4.8549 + 9.7098) × sin(52°) ≈ 39.0223 cm²

The area of the triangle ≈ 39.0223 cm²

(b) By sin rule, we have;

`\overline {RS}`/(sin(∠SQR)) =
`\overline {RQ}`/(sin(∠QSR))

By substituting, we have;

9.7098/(sin(∠SQR)) = 11.6799/(sin(97.11024))

sin(∠SQR) = 9.7098/(11.6799/(sin(97.11024))) ≈ 0.82493

∠SQR = sin⁻¹(0.82493) ≈ 55.582°.