1.三角形ABC中 AB=7 BC=3 CA=5 則 (1)sinB= 3/5 (2)sinB= 3/7
(3)cosB=5/7 (4)cosB=11/14 答案是4
運用餘弦定理(b^2)=(a^2)+(c^2)-2ac(cosB)
(5^2)=(7^2)+(3^2)-2(7)(3)cosB
25=49+9-42cosB
42cosB=33
cosB=33/42
=11/14
之後畫出直角三角形
斜邊AB線段長為14 , 底邊BC線段長為11
則運用畢氏定理解出對邊AC線段長
=根號(196-121)
=根號(75)
=5(根號3)
so sinB=[5(根號3)] / 14
故選(4)
2.如圖: AC=BC , AD:DC=3:2 ,則tanθ=? (3/7)
(圖在這邊:
http://www.flickr.com/photos/79991584@N05/7682783932/in/photostream)
設AD線段長為3r , CD線段長為2r
則BC線段長為3r+2r=5r
斜邊AB線段長為(5根號2)r [ 註:運用45度角1:1:(根號2) ]
BD線段長=根號[ (5r)^2+(2r)^2 ]
=根號(29r^2)
=(根號29)r
運用餘弦定理(b^2)=(a^2)+(d^2)-2ad(cosB)
(3r)^2=[ 5(根號2)r ]^2+[ (根號29)r ]^2-2[ (5根號2)r ][ (根號29)r ]cosθ
9(r^2)=50(r^2)+29(r^2)-10(根號58)(r^2)cosθ
10(根號58)(r^2)cosθ=70(r^2)
cosθ=[ 70(r^2) ] / [ 10(根號58)(r^2) ]
=70 / [10(根號58) ]
=7 / (根號58)
之後畫出直角三角形
斜邊長為(根號58) , 底邊長為7
則運用畢氏定理解出對邊長
=根號[ (根號58)^2-(7^2) ]
=根號(58-49)
=根號(9)
=3
so tanθ=3/7