(i)

|JK|² = 4² + 13²

|JK|² = 185

Take square root of both sides.

|JK|² = $\sqrt{\mathrm{185}}$ = 13.6 km ≈ 14 km (two significant figures)

(ii)

tan θ = $\frac{\mathrm{13\; km}}{\mathrm{4\; km}}$

θ = tan^-1($\frac{13}{4}$) = 72.9° ≈ 73°

Bearing of K from I + θ = 135°

Bearing of K from I = 135° - θ

Bearing of K from I = 135° - 73°

Bearing of K from I = 62°

Bearing of K from I is 062°

(i)

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c = 6$\sqrt{\mathrm{13}}$ since it's the side facing the 90° or the longest side

(6$\sqrt{\mathrm{13}}$)² = (2t)² + (3t)²

Note:
1. The square affects each terms in the bracket
2. Square root is cancelled out when squared

6² x 13 = 2² x t² + 3² x t²

6² x 13 = 4t² + 9t²

36 x 13 = 13t²

Divide both sides by 13

t² = 36

Take the square root of both sides.

t = $\sqrt{\mathrm{36}}$ = 6

(ii)

Bearing of M from K = 65° + θ°

You can apply the concept of SOHCAHTOA to solve for θ

We know the opposite and adjacent sides of θ hence tan θ is used

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan θ = $\frac{\mathrm{3t}}{\mathrm{2t}}$

tan θ = $\frac{3}{2}$

θ = tan^-1$\left(\frac{3}{2}\right)$

θ = 56.309932474 ≈ 56.31

Bearing of M from K = 65° + 56.31°

Bearing of M from K = 121.31° ≈ 121° (nearest whole number)

(α)

Apply cosine rule

|PR|² = 5² + 6² - 2 x 5 x 6 x cos 105

|PR|² = 25 + 36 + 14.46

|PR|² = 75.46

Take square root of both sides.

|PR| = $\sqrt{\mathrm{75.46}}$ = 8.68663012647 ≈ 8.67 km(two decimal places)

(β)

Bearing starts from the north.

The angle of bearing of R from P + θ = 60°

The angle of bearing of R from P = 60° - θ

Apply either cosine or sine rule to solve for θ

Method I

Using cosine rule

6² = 5² + 8.67² - 2 x 5 x 8.67 x cos θ

6² - 5² - 8.67² = - 2 x 5 x 8.67 x cos θ

-64.1689 = -86.7 x cos θ

Divide both sides by -86.7

cos θ = $\frac{-64.1689}{-86.7}$

cos θ = 0.74012572087

Take cos inverse

θ = cos^-1 (0.74012572087)

θ = 42.26°

The angle of bearing of R from P = 60° - 42.26°

The angle of bearing of R from P = 17.74°

Bearing of R from P = 17.75

Method II

Using the sine rule

$\frac{6}{\mathrm{sin\; \theta }}$ = $\frac{8.67}{\mathrm{sin\; 105}}$

Cross multiply

8.67 x sin θ = 6 x sin 105

sin θ = $\frac{\mathrm{6\; x\; sin\; 105}}{8.67}$

sin θ = 0.66846077943

Take sin inverse.

θ = sin^-1(0.66846077943)

θ = 41.9483778953 ≈ 41.9°

The angle of bearing of R from P = 60° - 41.9°

The angle of bearing of R from P = 18.05°

Bearing of R from P = 18.05°