Solution of Triangles Long Questions (Question 2)


Question 2:
The diagram shows a trapezium PQRS. PS is parallel to QR and QRS is obtuse. Find
(a) the length, in cm, of QS,
(b) the length, in cm, of RS,
(c) ÐQRS,
(d) the area, in cm2, of triangle QRS.

Solution:
(a)
QS sinP = PS sinQ QS sin 85 = 13.1 sin 28 QS= 13.1×sin 85 sin 28 QS=27.8 cm

(b)
ÐRQS = 180o – 85o – 28o
ÐRQS = 67o
Using cosine rule,
RS2 = QR2 + QS2 – 2 (QR)(QS) ÐRQS
RS2 = 6.42 + 27.82 – 2 (6.4)(27.8) cos 67o
RS2 = 813.8 – 139.04
RS2 = 674.76
RS = 25.98 cm

(c)
Using cosine rule, Q S 2 =Q R 2 +R S 2 2( QR )( RS )cosQRS 27.8 2 = 6.4 2 + 25.98 2 2( 6.4 )( 25.98 )cosQRS 772.84=715.92332.54cosQRS cosQRS= 715.92772.84 332.54 cosQRS=0.1712 QRS= 99.86

(d) Area of triangle QRS
= ½ (QR)(RS) sin R
= ½ (6.4) (25.98) sin 99.86o
= 81.91 cm2