Angle ABC
In a triangle A B C , D is the midpoint of B C . Join A D . Angle A D B = 4 5 degree and angle A C B = 3 0 degree. Find angle A B C
The answer is 105.
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4 solutions
Let ∣ C D ∣ = ∣ D B ∣ = x . Next, drop a perpendicular from A to the line extending through C B , letting the point of intersection be P . Now let ∣ A P ∣ = y and ∣ B P ∣ = z .
Then tan ( ∠ A C B ) = tan ( 3 0 ∘ ) = 2 x + z y ⟹ 2 x + z = 3 y and
tan ( ∠ A D B ) = tan ( 4 5 ∘ ) = x + z y ⟹ x + z = y .
So 2 ( x + z ) − ( 2 x + z ) = 2 y − 3 y ⟹ z = ( 2 − 3 ) y ⟹ z y = 2 − 3 1 = 2 + 3 .
But z y = tan ( ∠ A B P ) ⟹ ∠ A B P = tan − 1 ( 2 + 3 ) = 7 5 ∘ .
(Proof: tan ( 7 5 ∘ ) = tan ( 3 0 ∘ + 4 5 ∘ ) = 1 − tan ( 3 0 ∘ ) tan ( 4 5 ∘ ) tan ( 3 0 ∘ + tan ( 4 5 ∘ ) =
1 − 3 1 3 1 + 1 = 3 − 1 3 + 1 = 2 ( 3 + 1 ) 2 = 2 + 3 . )
Finally, we see that ∠ A B C = 1 8 0 ∘ − ∠ A B P = 1 8 0 ∘ − 7 5 ∘ = 1 0 5 ∘ .
I tried this problem twice, and both times came up with the
7
5
.
I created point P on
A
C
such that
D
P
is perpendicular to
A
C
.
I also designated DP to equal 10 units (I often do this on geometry problems looking for angles).
Then I simply chased the angles by trig ratios, law of cosines and law of sines. Here's what I got:
D
C
=
2
0
B
D
=
2
0
P
C
=
1
7
.
3
2
A
P
=
3
7
.
3
2
A
D
=
3
8
.
6
4
A
B
=
2
8
.
2
9
∠
A
B
C
=
7
5
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I like that approach. I get the same values except for the last, and I suspect that the fact that sin ( 7 5 ) = sin ( 1 0 5 ) may be the reason for the difference. Note that ∣ A C ∣ = ∣ A P ∣ + ∣ P C ∣ = 5 4 . 6 4 . Then ∣ A C ∣ cos ( 3 0 ) = 4 7 . 3 2 , which exceeds ∣ B C ∣ = ∣ D C ∣ + ∣ B D ∣ = 4 0 . This implies that ∠ A B C must be obtuse, as the perpendicular dropped from A will intersect the line extended from side B C at a point outside the triangle.
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After posting I walked away from it, and then yes, I did recall my unit circle and sin75 = sin 105. I'm guessing though that we sketched up the problem differently Brian. My sketch met all of the criteria, as did yours I assume. My method does not require <ABC to be obtuse. Recheck it. My math is solid. There just exists two answers to this question as posited. You can sketch it up with <ABC being acute and meeting the same initial criteria. Try it.
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@Ken Hodson – I'm sorry we're at odds here, but I just can't see how ∠ A B C can be acute. I did try again using some graph paper. I drew the base B C and placed D at the midpoint. I then drew a line at 3 0 ∘ above B C passing through C and a line 4 5 ∘ above B C passing through D , (both lines ascending from right to left), making their point of intersection the vertex A . Drawn this way, the projection of A to the horizontal lies beyond B , implying that ∠ A B C is obtuse. How does your sketch differ from this construction? I'm keeping an open mind, but I just can't see at the moment how else to construct the specified triangle.
From Tr. ADC, AC/AD= Sin(135°)/sin(30°) =√2 or AC²=2 AD². Further, by Apollonius, AB² + AC² = 2BD²+ 2AD². In other words, AB²=2BD²=BD BC. This implies, by the power point theorem, that AB is tangent to the circum-circle of Tr. ADC. Hence / BAD= / ACD = 30° (Alternate Segment Theorem) which implies that /_ABC=75°
Without loss of generality, let BC=2. S o B D = D C = 1 . ∠ C A D = 1 5 . ∠ A D C = 1 3 5 , ∠ D A B = 1 3 5 − B . Applying Sin Law to Δ A D B a n d Δ A D C , w e h a v e : − S i n D A B B D ∗ S i n B = A D = S i n C A D D C ∗ S i n 3 0 . S u b s t i t u t i n g t h e v a l u e s , S i n ( 1 3 5 − B ) 1 ∗ S i n ( B ) = S i n 1 5 1 ∗ S i n 3 0 = 2 ∗ C o s 1 5 . S i n ( B ) S i n ( 1 3 5 − B ) = 2 1 ∗ ( C o t B + 1 ) = 2 ∗ C o s 1 5 1 ⟹ C o t B = 2 ∗ C o s 1 5 1 − 1 . ∴ B = − 7 5 O R 1 0 5 o . B = 1 0 5 o