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RV-28-001 · Triangle area, unknown side and obtuse angle

In triangle ABC, AB = 17 cm, AC = 14 cm and angle BAC = 100°.
ABC17 cm14 cm100°
Diagram not accurately drawn

(a) Calculate BC and the area of ABC. Give your answers to 2 decimal places.

(b) Find the perpendicular distance from C to the line AB, to 2 decimal places.

Part (a) Cosine Rule ("Cosine Sausage") & Triangle Area (½ab sin C) Booklet p. 36 & p. 37
Method
1) Use Cosine Rule a² = b² + c² − 2bc cos(A) for BC
2) Use Area = ½ab sin(A)
3) Equate Area = ½ × AB × h to find perpendicular height h
Worked Example (RV-28-001 · Part (a))
Calculate BC and the area of ABC. Give your answers to 2 decimal places.
1)
The given angle is between the two known sides. Use cosine rule (p. 36): BC2=172+142−2(17)(14)cos⁡100∘BC^2=17^2+14^2-2(17)(14)\cos100^\circ.
2)
BC=485−(476)cos⁡100∘=23.83BC=\sqrt{485-(476)\cos100^\circ}=23.83 cm.
3)
Use the included-angle area formula (p. 37): Area=12(17)(14)sin⁡100∘=117.19\text{Area}=\frac12(17)(14)\sin100^\circ=117.19 cm².
Ans = BC = 23.83 cm; area = 117.19 cm².
Part (b) Cosine Rule ("Cosine Sausage") & Triangle Area (½ab sin C) Booklet p. 37
Method
1) Use ½ × base × perpendicular height for the same triangle
2) Rearrange to find its perpendicular height
Worked Example (RV-28-001 · Part (b))
Find the perpendicular distance from C to the line AB, to 2 decimal places.
1)
Use the same area with AB as the base: 12(17)h=12(17)(14)sin⁡100∘\frac12(17)h=\frac12(17)(14)\sin100^\circ.
2)
Multiply by 2 and divide by 17: h=14sin⁡100∘=13.79h=14\sin100^\circ=13.79 cm. This is the perpendicular distance to the line AB.
Ans = 13.79 cm