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maths question
1.cos30度xcos45度+sin60度xsin45度 2. cos^2 45度xsin30度 _________________ tan^2 60度 3. tan60度 = 開方8xcos45度 _______ sin(acute angle/3+35度) 4.BCDE is a triangle and ABC is a straight line. If Ad is 20cm,AE =10x開方2and ADE=60度,find the lenght of CD and cx
最佳解答:
(I omitted the degree notation) sqrt. denotes square root. (1) (cos 30)(cos 45)+(sin 60)(sin 45) =(2 sqrt.)/2[2 cos 30] =(2 sqrt.)cos 30 =(2 sqrt.)(3 sqrt./2) =6 sqrt./2 (2) (cos245)(sin 30)/tan260 =(2 sqrt./2)2(1/2)/(3 sqrt.)2 =(1/2)(1/2)/3 =1/12 (3) tan 60/sin (θ/3+35)=8 sqrt. cos 45 3 sqrt./sin (θ/3+35)=2(2 sqrt.)(2 sqrt.)/2 3 sqrt./sin (θ/3+35)=2 sin (θ/3+35)=3 sqrt./2 Because θ is an acute angle, therefore there is only one solution for θ/3+35 θ/3+35=60 θ/3=25 θ=75 (4) Is there a figure? I think it is not showing clearly.
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maths question
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發問:1.cos30度xcos45度+sin60度xsin45度 2. cos^2 45度xsin30度 _________________ tan^2 60度 3. tan60度 = 開方8xcos45度 _______ sin(acute angle/3+35度) 4.BCDE is a triangle and ABC is a straight line. If Ad is 20cm,AE =10x開方2and ADE=60度,find the lenght of CD and cx
最佳解答:
(I omitted the degree notation) sqrt. denotes square root. (1) (cos 30)(cos 45)+(sin 60)(sin 45) =(2 sqrt.)/2[2 cos 30] =(2 sqrt.)cos 30 =(2 sqrt.)(3 sqrt./2) =6 sqrt./2 (2) (cos245)(sin 30)/tan260 =(2 sqrt./2)2(1/2)/(3 sqrt.)2 =(1/2)(1/2)/3 =1/12 (3) tan 60/sin (θ/3+35)=8 sqrt. cos 45 3 sqrt./sin (θ/3+35)=2(2 sqrt.)(2 sqrt.)/2 3 sqrt./sin (θ/3+35)=2 sin (θ/3+35)=3 sqrt./2 Because θ is an acute angle, therefore there is only one solution for θ/3+35 θ/3+35=60 θ/3=25 θ=75 (4) Is there a figure? I think it is not showing clearly.
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