Let quadrilateral ABCD satisfy angle BAC= angle CAD=2 angle ACD=40^circ and angle ACB=70^circ. Find angle ADB.

Ximena Skinner 2022-07-15 Answered
Let quadrilateral ABCD satisfy B A C = C A D = 2 A C D = 40 and A C B = 70 . Find A D B.
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Answers (2)

Dalton Lester
Answered 2022-07-16 Author has 12 answers
Step 1
77 , 34 , so

Step 2
sin 80 sin 40 sin ( 70 x ) sin x sin 20 sin 90 = 1
After some manipulation we get cot x = tan 20 + 2 cos 10 x = . . .
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Joel French
Answered 2022-07-17 Author has 10 answers

Since A C B = A B C = 70 , the triangle ABC is isosceles and A B ¯ = A C ¯ .
By applying the law of sines to the triangle ACD, we get that:
A D ¯ = A C ¯ sin A C D sin A D C = A C ¯ sin 20 sin 120 = 2 A C ¯ sin 20 3 .
And, by applying the law of sines to the triangle ABD, we get that:
A D ¯ sin A D B = A B ¯ sin A B D . ( )
Let α = A D B ..
Step 2
Since A D ¯ = 2 A C ¯ sin 20 3 , A B ¯ = A C ¯ and A B D = 100 α , , the equality (∗) turns into:
2 A C ¯ sin 20 sin α 3 = A C ¯ sin ( 100 α ) ,
2 sin 20 sin α = 3 sin ( 90 + 10 α ) ,
4 sin 10 cos 10 sin α = 3 cos ( 10 α ) ,
4 sin 10 cos 10 sin α = 3 ( cos 10 cos α + sin 10 sin α ) ,
4 sin 10 sin α = 3 ( cos α + tan 10 sin α ) ,
( 4 sin 10 3 tan 10 ) sin α = 3 cos α ,
tan α = 3 4 sin 10 3 tan 10 .
Hence, A D B = α = arctan ( 3 4 sin 10 3 tan 10 ) 77 , 3361794 .
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