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In △ ABC, ab = AC = 10, BC = 16, then CoSb = (), tanb = ()
In isosceles △ ABC, ab = AC = 10, BC = 16,
If a is used as ad ⊥ BC over D, then BD = 8,
In RT △ abd, ab = 10, BD = 8, then
AD=√AB²-BD²=√10²-8²=6,
cosB=BD/AB=8/10=4/5
So tanb = ad / BD = 6 / 8 = 3 / 4
1. In △ ABC, ab = AC = 10, BC = 16, then tanb = opposite edge to leading edge = 10 to 10 = 1______ .
No, sit high to BC through point a, and hand BC to d. ∵ AB = AC = 10, BC = 16 ∵ BD = CD = 1 / 2, BC = 8 ∵ ADB = 90 螧 = 10, BD = 8. According to Pythagorean theorem, ad = √ (AB?; D? Tanb = ad: BD = 6:8 = 3 / 4 = 0.75, hope to adopt, thank you!
In △ ABC, ab = AC, BC = 4, Tan B = 2
(1) Finding the length of ab
(2) Point D is on BC, point E is on AC, ∠ ade = ∠ B. If △ ade is similar to △ abd, find the length of CD
(3) If △ ade is an isosceles triangle, find the length of CD
(1) AB = 2 times the root 5, as ad vertical BC will get the answer
(2) CD = 2, actually point D is the midpoint of BC
In angle ABC, ab = AC = 5, BC = 6, then what is tanb?
cosB=(AB^2+BC^2-AC^2)/(2AB*BC)=(36)/(2*5*6)=36/60=3/5 -->sinB>0 -->sinB^2=1-cosb^2 -->sinB=4/5 -->tanB=sinB/cosB=4/3
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