triangle abc d is the midpoint of bc (angle)ADB=45 degree , (angle)ACD=30 degree find (angle)BAD

Question

(angle)ACD=30 degree  find the angle measure of (angle)BAD

Answer 1

15 degree
triangle property: Angles opposite to equal sides are equal

Answer 2

1. To solve this problem, use the trigonometry/ specific angles value table that gives the values in radical forms, or the calculator equiped with trigonometric functions.
2. Let length of BD and DC be BD=DC=a.  In △ADC, ∠ADC=135° and ∠CAD=15°  By the law of sines, AC/sin135° = DC/sin15° = AD/sin30°, so AC = DCsin135°/sin15°  Use identity, sin135° = sin(180°－45°) =sin45°  Thus AC is written and simplified as follows: AC = a*sin45°/sin15° = a*(√2/2)/(√6－√2)/4,  so AC = (1+√3)a
3. Draw an altitude from A to BC, and label the foot H.  HC = ACcos30° = (1+√3)a(√3/2) = (3+√3)a/2  So BH is written and simplified as follows: BH =HC－BC = (3+√3)a/2－2a,  Thus BH = (√3－1)a/2  Since BH is possitive, therefore foot H is on the extention of BC.
4. In △ABH, AH = ACsin30° = (1+√3)a/2   So tan∠ABH = AH/BH = (1+√3)a/2 / (√3－1)a/2 = 2+√3  Thus ∠ABH=arc tan(2+√3)=75°  and ∠ABD=105°, ∠BAD=30° and ∠BAC=45°
5. CK: In △ABD, by the law of sines, AB = BDsin45°/sin30° = a√2   In triangle ABC, AB/sin30° = BC/sin45° = AC/sin105° = 2a√2   CKD.  Therefore required angle BAD is 30 degree.

Answer 3

1. Draw△ABC on a graph.  Mark points: B at the origin,(0,0), D at pt.(1,0) and C at pt.(2,0).  so BD=DC=1  Pt.A will be marked above the x-axis.
2. Draw 2 lines, y1 and y2: line y1 corresponds to AD passing thru D(1,0) with the slope,-1 or        -tan45°, and line y2 corresponds to AC passing thru C(2,0) with the slope, -√3/3 or -tan30°.  Pt.A is the intersection of the 2 lines.　The equation of each line will be defined as follows:   y1=-x+1, y2=-√3/3x+2√3/3.  Solve these equations.  x=(1-√3)/2 and y1=y2=(1+√3)/2               The cordinates of A is A{(1-√3)/2, (1+√3)/2}.
3. Connect A to B, and draw an altitude from A to x-axis intersecting x-axis at H{(1-√3)/2, 0}.    Put ∠ABH=θ   tanθ =AH/|BH| = {(1+√3)/2} / {(√3-1)/2} = 2+√3 ***
4. In △ABD, let ∠BAD, ∠ABD and ∠ADB be α, β and δ respectively.  By a formula of tangent on a triangle,  tanα+tanβ+tanδ = tanα*tanβ*tanδ   Here, tanδ = tan45°=1 and tanβ = tan(180-θ)= -tanθ = -(2+√3).   The equation shown above will be rewritten and simplified as follows: tanα+ tanβ+1=tanα*tanβ, so tanα=(tanβ+1)/(tanβ-1) and tanβ=-(2+√3).   Thus tanα= {-(2+√3)+1} / {-(2+√3)-1}=1/√3, so α=∠BAD=30°    Therefore the required angle BAD is 30 degree.  *** From here, if you use the angle value table or calculator, you can get the angle asked.        　　 

Answer 4

Sw=p-81

tv=p-90