In Δ A B C , if the length of B C is twice the length of A C , and ∠ A − ∠ B = 9 0 ∘ , what is the value of tan C ?
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By the Sine Rule , a = 2 b ⇒ sin A = 2 sin B .
Now, A = 2 π + B
∴ sin ( 2 π + B ) = cos B
cos B = 2 sin B
tan B = 2 1
Similarly, B = A − 2 π
∴ sin A = sin ( A − 2 π )
sin A = − 2 cos A
tan A = − 2
Now, we know that tan A + tan B + tan C = tan A tan B tan C
Substituting these values, we get tan C = 4 3
Wow!You'd definitely laugh at me if you knew what misery I've gone through.Nice solution!
In the third line, I think is cos B instead of sin B . Nice solution!
Alternately, C = Pi - (A+B) and Tan C = - Tan (A + B) =(Tan A + Tan B)/((Tan A)(Tan B) - 1) = (-2+(1/2))/(-2(1/2)-1) = 3/4
Let B D = 2 A C .
Using the law of sines ⟹ s i n B s i n A = 2 and s i n A s i n B = 2 1 ⟹ s i n ( A ) = 2 s i n B and s i n B = 2 1 s i n A
∠ A − ∠ B = 9 0 ⟹
s i n A = s i n ( 9 0 + B ) = c o s B
s i n B = s i n ( A − 9 0 ) = − c o s A
⟹ s i n A = − 2 c o s A ⟹ t a n A = − 2
s i n B = 2 1 c o s B ⟹ t a n B = 2 1
⟹
t a n C = t a n ( 1 8 0 − ( A + B ) ) = − t a n ( A + B ) = − ( 1 − t a n A t a n B t a n A + t a n B ) = 4 3 .
It is given that A − B = 9 0 ∘ ⟹ A = 9 0 ∘ + B .
⟹ C = 1 8 0 ∘ − A − B = 1 8 0 ∘ − ( 9 0 ∘ + B ) − B = 9 0 ∘ − 2 B
Using sine rule, we have:
2 sin A ⟹ sin A sin ( 9 0 ∘ + B ) ⟹ tan B = 1 sin B = 2 sin B = 2 sin B = 2 1
Now, we have:
tan C = tan ( 9 0 ∘ + 2 B ) = tan ( 2 ( 4 5 ∘ − B ) ) tan ( 4 5 ∘ − B ) = 1 + tan B 1 − tan B = 1 + 2 1 1 − 2 1 = 3 1 = 1 − ( 3 1 ) 2 2 ( 3 1 ) = 8 6 = 4 3 = 0 . 7 5
given that , BC = 2AC
=> BC/AC = 2
denoting AB,BC , CA by c,b,a, respectively , we can now write a/b = 2
as we know from Sine law , a/SinA = b/SinB ,
we get , a/b = SinA/SinB = 2 ... Let it be eqn no (1)
again , given that, A-B = pi/2
or, A = pi/2+B
so , eqn (1) now becomes ,
Sin(pi/2+B)/SinB = 2
but Sin(pi/2+X) = CosX ,
so , from (1) , CosB/SinB = 2
it implies, CotB = 2
or,tanB =1/2
or, B = arctan(1/2) = 26.56505118 degree (aproximately)
again , for any triangle ,
A+ B + C = pi
or, pi/2 + B + B +C =pi
or, 2B + C = pi/2
or, C = pi/2 - 2B
putting the value of B , we will get C = 36.86989765(aproximately)
now tanC = 3/4
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As a = 2 b ⇒ sin A = 2 sin B but as A = B + 2 π , sin A = cos B
cos B = 2 sin B ⇒ tan B = 2 1
Then C = π − B − B − 2 π = 2 π − 2 B and by trigonometric identities tan C = tan ( 2 π − 2 B ) = cot 2 B .
tan C = cot 2 B = tan 2 B 1 = 2 sin B cos B cos 2 B − sin 2 B
Drawing a triangle we get:
sin B = 5 1
cos B = 5 2
∴
tan C = 2 ( 5 1 ) ( 5 2 ) ( 5 2 ) 2 − ( 5 1 ) 2 = 4 3