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F.4 Maths trigonometry
發問:
1.PQR is an isosceles triangular metal sheet and M is the mid-point of QR. PQ=PR=5cm and QR=6cm. The metal sheet is folded along PM such that ∠QMR=50° and placed on a horizontal surface. P,Q and R lie on the same horizontal surface.a)find the distance between Q and R on the horizontal... 顯示更多 1.PQR is an isosceles triangular metal sheet and M is the mid-point of QR. PQ=PR=5cm and QR=6cm. The metal sheet is folded along PM such that ∠QMR=50° and placed on a horizontal surface. P,Q and R lie on the same horizontal surface. a)find the distance between Q and R on the horizontal surface b)find the areas of triangle MQR and triangle PQR c)find the shortest distance from M to the horizontal surface d)find the angle between the plane MQR and the horizontal surface e)describe how the volume of the tetrahedron PQRM changes when ∠QMR increases from 50° to 120°. explain your answer 上傳不了圖片...如果真的不明白請留下e-mail,我會把圖片send去,謝謝! 更新: RE:YA HOO ! 知識+管理員 是直接把「連接」那個網址copy在這裏嗎？ 但是我之前這樣做被説違規和扣點數........ 更新 2: p.s.已在意見貼上相片，thx
你可以到 http://postimage.org/ 上傳圖片 2014-05-03 13:57:45 補充： 咁就唔好COPY啦~ 話說扣點數呢家野真係... 2014-05-03 15:03:16 補充： a) Since M is the mid-pt of QR MQ=MR=3 cm Also, we know that ∠QMR=50°. So, by the cosine formula, QR^2 = MQ^2 + MR^2 - 2(MQ)(MR) cos∠QMR QR^2 = 3^2 + 3^2 - 2(3)(3) cos 50° QR^2 = 18 - 18 cos 50° QR = 2.53570957 cm QR = 2.54 cm ( cor to 3 sig fig ) b) For the triangle MQR, we know that MQ=MR=3 cm , ∠QMR=50° Therefore, The area of triangle MQR =(1/2)(MQ)(MR) sin∠QMR =(1/2)(3)(3) sin 50° =(9/2) sin 50° =3.45 cm^2 ( cor to 3 sig fig ) For the triangle PQR, we know that PQ=PR=5 cm , QR=2.53570957 cm s = (5+5+2.53570957)/2 = 6.267854785 cm By the Heron's formula, The area of triangle PQR =√[ (6.267854785)(6.267854785-5)(6.267854785-5)(6.267854785-2.53570957) ] =6.132086096 cm =6.13 cm^2 ( cor to 3 sig fig ) c) Now, we know the areas of triangle PQR and triangle MQR. Let h be the shortest distance from M to the horizontal surface. PM^2 =PR^2 - MR^2 PM = √(5^2 - 3^2) = √16 = 4 cm For the volume of tetrahedron MPQR, volume =(1/3)(Area of triangle PQR)(h) ... 1 Also, the volume = (1/3)(Area of triangle MQR)(PM) ... 2 Combine 1 and 2, (1/3)(Area of triangle PQR)(h) = (1/3)(Area of triangle MQR)(PM) (Area of triangle PQR)(h) = (Area of triangle MQR)(PM) 6.132086096 h = [(9/2) sin 50°] (4) h = 2.248631177 cm h = 2.25 cm ( cor to 3 sig fig ) Therefore, the shortest distance from M to the horizontal surface is 2.25 cm. d) Let N be the mid-pt of QR. In triangle MNR and triangle MNQ, MR=MQ ... given MN=MN ... common QN=RN Therefore, triangle MNR ~= triangle MNQ. ( SSS ) So ∠NMQ=∠NMR=50°/2=25° ... ( corr ∠s, ~= tiangles ) MN=MR cos 25° MN = 2.718923361 cm Let θ be the angle between the plane MQR and the horizontal surface. sin θ = h / MN sin θ = 2.248631177 / 2.718923361 θ=55.8° ( cor to 3 sig fig ) Therefore, the angle between the plane MQR and the horizontal surface is 55.8°. e) 因為字數問題，請到意見區看看。 2014-05-03 15:03:42 補充： e) When ∠QMR increases from 50° to 90°, the area of triangle MRQ will increase. Then the volume will also increase because volume = (1/3)(Area of triangle MRQ)(PM) where PM is fixed. When ∠QMR increases from 90° to 120°, the area of triangle MRQ will decrease. Then the volume will also decrease.
其他解答:
(a) QR^2=3^2+3^2-(3)(2)cos50degree QR=2.54cm (to 3 s.f.) (b) areas of triangle MQR=1/2*3*3*sin50 =3.45cm^2 (to 3 s.f.) s=1/2*(5+5+6)=8cm so, Area of triangle PQR=√8*(8-5)*(8-5)*(8-6) =12cm^2 (c) consider triangle MQR (from (b)) and QR (from (a)) 1/2*3*3*sin50=1/2*(√3^2+3^2-(3)(2)cos50degree)*required shortest distance required shortest distance=2.716535...cm (d) ******************************************************************************************** (e) ********************************************************************************************|||||貼在這裏應該可以吧0_0 http://postimg.org/image/cwod5p6vn/
F.4 Maths trigonometry
發問:
1.PQR is an isosceles triangular metal sheet and M is the mid-point of QR. PQ=PR=5cm and QR=6cm. The metal sheet is folded along PM such that ∠QMR=50° and placed on a horizontal surface. P,Q and R lie on the same horizontal surface.a)find the distance between Q and R on the horizontal... 顯示更多 1.PQR is an isosceles triangular metal sheet and M is the mid-point of QR. PQ=PR=5cm and QR=6cm. The metal sheet is folded along PM such that ∠QMR=50° and placed on a horizontal surface. P,Q and R lie on the same horizontal surface. a)find the distance between Q and R on the horizontal surface b)find the areas of triangle MQR and triangle PQR c)find the shortest distance from M to the horizontal surface d)find the angle between the plane MQR and the horizontal surface e)describe how the volume of the tetrahedron PQRM changes when ∠QMR increases from 50° to 120°. explain your answer 上傳不了圖片...如果真的不明白請留下e-mail,我會把圖片send去,謝謝! 更新: RE:YA HOO ! 知識+管理員 是直接把「連接」那個網址copy在這裏嗎？ 但是我之前這樣做被説違規和扣點數........ 更新 2: p.s.已在意見貼上相片，thx
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最佳解答:你可以到 http://postimage.org/ 上傳圖片 2014-05-03 13:57:45 補充： 咁就唔好COPY啦~ 話說扣點數呢家野真係... 2014-05-03 15:03:16 補充： a) Since M is the mid-pt of QR MQ=MR=3 cm Also, we know that ∠QMR=50°. So, by the cosine formula, QR^2 = MQ^2 + MR^2 - 2(MQ)(MR) cos∠QMR QR^2 = 3^2 + 3^2 - 2(3)(3) cos 50° QR^2 = 18 - 18 cos 50° QR = 2.53570957 cm QR = 2.54 cm ( cor to 3 sig fig ) b) For the triangle MQR, we know that MQ=MR=3 cm , ∠QMR=50° Therefore, The area of triangle MQR =(1/2)(MQ)(MR) sin∠QMR =(1/2)(3)(3) sin 50° =(9/2) sin 50° =3.45 cm^2 ( cor to 3 sig fig ) For the triangle PQR, we know that PQ=PR=5 cm , QR=2.53570957 cm s = (5+5+2.53570957)/2 = 6.267854785 cm By the Heron's formula, The area of triangle PQR =√[ (6.267854785)(6.267854785-5)(6.267854785-5)(6.267854785-2.53570957) ] =6.132086096 cm =6.13 cm^2 ( cor to 3 sig fig ) c) Now, we know the areas of triangle PQR and triangle MQR. Let h be the shortest distance from M to the horizontal surface. PM^2 =PR^2 - MR^2 PM = √(5^2 - 3^2) = √16 = 4 cm For the volume of tetrahedron MPQR, volume =(1/3)(Area of triangle PQR)(h) ... 1 Also, the volume = (1/3)(Area of triangle MQR)(PM) ... 2 Combine 1 and 2, (1/3)(Area of triangle PQR)(h) = (1/3)(Area of triangle MQR)(PM) (Area of triangle PQR)(h) = (Area of triangle MQR)(PM) 6.132086096 h = [(9/2) sin 50°] (4) h = 2.248631177 cm h = 2.25 cm ( cor to 3 sig fig ) Therefore, the shortest distance from M to the horizontal surface is 2.25 cm. d) Let N be the mid-pt of QR. In triangle MNR and triangle MNQ, MR=MQ ... given MN=MN ... common QN=RN Therefore, triangle MNR ~= triangle MNQ. ( SSS ) So ∠NMQ=∠NMR=50°/2=25° ... ( corr ∠s, ~= tiangles ) MN=MR cos 25° MN = 2.718923361 cm Let θ be the angle between the plane MQR and the horizontal surface. sin θ = h / MN sin θ = 2.248631177 / 2.718923361 θ=55.8° ( cor to 3 sig fig ) Therefore, the angle between the plane MQR and the horizontal surface is 55.8°. e) 因為字數問題，請到意見區看看。 2014-05-03 15:03:42 補充： e) When ∠QMR increases from 50° to 90°, the area of triangle MRQ will increase. Then the volume will also increase because volume = (1/3)(Area of triangle MRQ)(PM) where PM is fixed. When ∠QMR increases from 90° to 120°, the area of triangle MRQ will decrease. Then the volume will also decrease.
其他解答:
(a) QR^2=3^2+3^2-(3)(2)cos50degree QR=2.54cm (to 3 s.f.) (b) areas of triangle MQR=1/2*3*3*sin50 =3.45cm^2 (to 3 s.f.) s=1/2*(5+5+6)=8cm so, Area of triangle PQR=√8*(8-5)*(8-5)*(8-6) =12cm^2 (c) consider triangle MQR (from (b)) and QR (from (a)) 1/2*3*3*sin50=1/2*(√3^2+3^2-(3)(2)cos50degree)*required shortest distance required shortest distance=2.716535...cm (d) ******************************************************************************************** (e) ********************************************************************************************|||||貼在這裏應該可以吧0_0 http://postimg.org/image/cwod5p6vn/
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