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In RT △ ABC, ∠ C = 90 °, ab = 8, AC = 4 times the root sign 3, find the degree of ∠ a, ∠ B
sinB=ACAB=4√3/8=√3/2,
∴∠B=60°,
∴∠A=90°-60°=30°.
As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 2, radical 3, ab = 3, root 2, calculate the perimeter and area of triangle ABC
According to Pythagorean theorem, BC = root 6, so area s = root 6 x 2 root sign 3 times 0.5 = 3 root sign 2
Perimeter = 2 roots 3 + 3 roots 2 + 6 roots
Known triangle ABC, angle B = 45 degrees, AC = root 10, COSC = 2 / 5, root sign 5, find BC ∵ COSC = (2 roots 5) / 5 > 0 C is an acute angle Now, I use AC / sin45 = AB / sinc to calculate AB = 2, and then use COSC = a ^ 2 + B ^ 2-C ^ 2 / 2Ab to figure out why BC has two values. One is root 2 and the other is root 2. My answer is 3 root sign 2. What's wrong? Why is this method wrong?
This method is right, but you need to exclude a solution. You have proved that C is an acute angle, so from sinc = √ 5 / 5, C is less than 45 degrees. So a must be an obtuse angle, so take the larger value of the solution, which is 3 √ 2
Triangle ABC. Angle B is equal to 45 degrees, AC is equal to root 10, COSC is equal to two fifths of 5, root sign 5, find Sina and ab
cosC=2/5√5,SinC=√[1-(2/5√5)^2]=3/5,AB/SinC=AC/SinB,
AB=AC/SinB*SinC=√10/Sin45°*3/5=6/5√5
sinA=sin(∏-45°-C)=-sin(45°+C)=-Sin45°*cosC-cos45°SinC=-√10/5-3/10√2
In the triangle ABC, the angle B = 45 ° AC = root 10, COSC = (root 20) / 5 BC =? Well, we don't want to do vertical. We can't do vertical. Add up the two paragraphs. Er, we have to get the result directly
sinC=√5/5
AB/sinC=AC/sinB
AB=2
cosB=√2/2=(2^2+BC^2-10)/(2*2*BC)
BC=3√2
In the triangle ABC, ab = 5 under the root sign, AC = 5, cosc9 / 10, then the value of BC is?
Cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2abcosca = BC, B = AC, C = abcosc = (a ^ 2 + B ^ 2-C ^ 2) / 2ab9 / 10 = (a ^ 2 + 25-5) / 2A * 510a ^ 2 + 200-90a = 0A ^ 2-9a + 20 = 0 (A-4) (a-5) = 0A = 4, a = 5BC = 4 or 5. If you don't understand this question, you can ask me if you don't understand it, please post and click my picture for help
In the triangle ABC, if AB = radical 5, AC = 5, and COSC = 9 / 10, then BC =?
Cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2abcosc
a=BC,b=AC,c=AB
cosC=(a^2+b^2-c^2)/2ab
9/10=(a^2+25-5)/2a*5
10a^2+200-90a=0
a^2-9a+20=0
(a-4)(a-5)=0
a=4,a=5
BC = 4 or 5
In the triangle ABC, B = 45, AC = root 10, COSC = 2 / 5, Radix 5. (1) find BC. (2) if D is the midpoint of AB, find the length of the center line CD (1) Find BC (2) If D is the midpoint of AB, find the length of the center line CD
(1) Make AE perpendicular to BC
Right triangle AEC
COSC = EC / AC, AC = root 10, COSC = 2 / 5, root number 5
EC = 2 root sign 2
AE = root (AC square - EC Square) = root 2
Right triangle Abe
AE = 2, be = 2
BC = be + CE = 3 root number 2
(2) DF is made perpendicular to BC
In the right triangle Abe, D is the midpoint of ab
AE=2DF
DF = 1 / 2 root number 2
B = 45, BF = DF = 1 / 2 root number 2
Right triangle AEC
FC = bc-bf = 3 root number 2-1 / 2 root number 2 = 5 / 2 root number 2
DC = root (DF square + CF Square) = root 13
The picture should be drawn by yourself
AB in △ ABC= 5，AC=5，cosC=9 10, then the value of BC is () A. 4 B. 5 C. 4 or 5 D. 2 or Five
∵AB=
5，AC=5，cosC=9
10，
According to the cosine theorem, AB2 = ac2 + bc2-2ac · bccosc
∴5=25+BC2-2×5×BC×9
Ten
Bc2-9bc + 20 = 0
It can be obtained that BC = 4 or BC = 5
Therefore, C is selected
AB in △ ABC= 5，AC=5，cosC=9 10, then the value of BC is () A. 4 B. 5 C. 4 or 5 D. 2 or Five
∵AB=
5，AC=5，cosC=9
10，
According to the cosine theorem, AB2 = ac2 + bc2-2ac · bccosc
∴5=25+BC2-2×5×BC×9
Ten
Bc2-9bc + 20 = 0
It can be obtained that BC = 4 or BC = 5
Therefore, C is selected
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