Question

The diagram shows five straight roads.
PQ = 4.5 km, QR = 4 km and PR = 7 km.
Angle RPS = 40° and angle PSR = 85º.
(a) Calculate angle PQR and show that it rounds to 110.7º.

[4]
(b) Calculate the length of the road RS and show that it rounds to 4.52 km.

c)Calculate the area of the quadrilateral PQRS


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Answer 1

Answer:

B calculate the length of the road rs and show that it rounds to 4.52 km.

Answer 2

Answer:(a)

Step-by-step explanation: (use cos rule)

cos A= $\frac{b^2 + c^2 - c^2}{2\times bc }$

cos A= $\frac{4.5^2 + 4^2 - 7^2}{2\times 4.5\times4 }$

cos A= 110.74

cos A = 110.7

Answer:(b)

Step-by-step explanation: (use sin rule)

sin rule = $\frac{a}{sinA}$ = $\frac{b}{sinB}$

=  $\frac{a}{sin40}$ = $\frac{7}{sin85\\}$

= a × sin 85 = sin40 × 7

a = $\frac{sin 40 \times 7}{sin 85}$

= 4.516

=4.52

Answer:(c)

Step-by-step explanation:  

Area of triangle = $\frac{1}{2}$ × a × b× sin c

180- 85-40 = 55

triangle 1 = $\frac{1}{2\\}$ × 7 × 4.52 × sin 55

= 12.95

triangle 2 = $\frac{1}{2\\}$ × 4 × 4.55 × sin 110.7

= 8.4

area of quadrilateral PQRS = 12.95 + 8.4

=21.35