Form 4 Trigonometry Questions
2017/07/04 15:13
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Form 4 Trigonometry Questions
發問:
AB is a vertical pole of height 4m. C is a point 10m due east of B and AC is a tightened rope. The sun shines from N30°W at an angle of elevation 60° and the shadow of AB and AC on the level ground are FB and FC respectively. Find (a)FB and FC, (b)∠BFC.
最佳解答:
Angle AFB = elevation of the sun = 60 degree. From triangle AFB, AB/FB = tan 60, that is 4/FB = tan 60, FB = 4/tan 60 = 2.309 m. Since the sun is N30W, so angle FBC = 90 - 30 = 60 degree. By cosine rule, FC^2 = FB^2 + BC^2 - 2(FB)(BC) cos 60 = (4/tan 60)^2 + (10)^2 - 2(4/tan 60)(10) cos 60 = 5.33333 + 100 - 23.094 = 82.23932 so FC = 9.0686 m. By sine rule, from triangle BFC, BC/sin(angle BFC) = FC/sin 60 angle BFC = arcsin [BC sin 60/FC] = arcsin [ 8.660254/9.0686] = 72.7 degree.
Form 4 Trigonometry Questions
發問:
AB is a vertical pole of height 4m. C is a point 10m due east of B and AC is a tightened rope. The sun shines from N30°W at an angle of elevation 60° and the shadow of AB and AC on the level ground are FB and FC respectively. Find (a)FB and FC, (b)∠BFC.
最佳解答:
Angle AFB = elevation of the sun = 60 degree. From triangle AFB, AB/FB = tan 60, that is 4/FB = tan 60, FB = 4/tan 60 = 2.309 m. Since the sun is N30W, so angle FBC = 90 - 30 = 60 degree. By cosine rule, FC^2 = FB^2 + BC^2 - 2(FB)(BC) cos 60 = (4/tan 60)^2 + (10)^2 - 2(4/tan 60)(10) cos 60 = 5.33333 + 100 - 23.094 = 82.23932 so FC = 9.0686 m. By sine rule, from triangle BFC, BC/sin(angle BFC) = FC/sin 60 angle BFC = arcsin [BC sin 60/FC] = arcsin [ 8.660254/9.0686] = 72.7 degree.
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