Trigonometry!

Geometry Level 2

In Δ A B C \Delta ABC , if the length of B C BC is twice the length of A C , AC, and A B = 9 0 , \angle A-\angle B=90^\circ, what is the value of tan C \tan C ?


The answer is 0.75.

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5 solutions

Paola Ramírez
May 24, 2015

As a = 2 b sin A = 2 sin B a=2b \Rightarrow \sin A=2\sin B but as A = B + π 2 , sin A = cos B A=B+\frac{\pi}{2}, \sin A=\cos B

cos B = 2 sin B tan B = 1 2 \cos B= 2 \sin B \Rightarrow \tan B=\frac{1}{2}

Then C = π B B π 2 = π 2 2 B C=\pi-B-B-\frac{\pi}{2}=\frac{\pi}{2}-2B and by trigonometric identities tan C = tan ( π 2 2 B ) = cot 2 B \tan C=\tan(\frac{\pi}{2}-2B)=\cot 2B .

tan C = cot 2 B = 1 tan 2 B = cos 2 B sin 2 B 2 sin B cos B \tan C=\cot 2B=\frac{1}{\tan 2B}=\frac{\cos^2 B- \sin^2 B}{2\sin {B}\cos {B}}

Drawing a triangle we get:

sin B = 1 5 \sin B=\frac{1}{\sqrt{5}}

cos B = 2 5 \cos B=\frac{2}{\sqrt{5}}

\therefore

tan C = ( 2 5 ) 2 ( 1 5 ) 2 2 ( 1 5 ) ( 2 5 ) = 3 4 \tan C=\frac{(\frac{2}{\sqrt{5}})^2- (\frac{1}{\sqrt{5}})^2}{2 (\frac{1}{\sqrt{5}})(\frac{2}{\sqrt{5}})}=\boxed{\frac{3}{4}}

Omkar Kulkarni
Feb 19, 2015

By the Sine Rule , a = 2 b sin A = 2 sin B . a=2b\Rightarrow\sin A=2\sin B.

Now, A = π 2 + B A=\frac{\pi}{2}+B

sin ( π 2 + B ) = cos B \therefore \sin \left(\frac{\pi}{2}+B\right)=\cos B

cos B = 2 sin B \cos B=2\sin B

tan B = 1 2 \tan B = \frac{1}{2}

Similarly, B = A π 2 B=A-\frac{\pi}{2}

sin A = sin ( A π 2 ) \therefore \sin A = \sin \left(A-\frac{\pi}{2}\right)

sin A = 2 cos A \sin A = -2\cos A

tan A = 2 \tan A = -2

Now, we know that tan A + tan B + tan C = tan A tan B tan C \tan A+\tan B+\tan C = \tan A\tan B\tan C

Substituting these values, we get tan C = 3 4 \boxed{\tan C = \frac{3}{4}}

Wow!You'd definitely laugh at me if you knew what misery I've gone through.Nice solution!

Arian Tashakkor - 6 years ago

In the third line, I think is cos B \cos B instead of sin B \sin B . Nice solution!

Paola Ramírez - 6 years ago

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Yup. Thanks!

Omkar Kulkarni - 6 years ago

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

Jeganathan Sriskandarajah - 5 years, 4 months ago
Rocco Dalto
Oct 1, 2016

Let B D = 2 A C . {\bf BD = 2AC. }

Using the law of sines s i n A s i n B = 2 {\bf \implies \frac{sinA}{sinB} = 2 } and s i n B s i n A = 1 2 s i n ( A ) = 2 s i n B {\bf \frac{sinB}{sinA} = \frac{1}{2} \implies sin(A) = 2sinB } and s i n B = 1 2 s i n A {\bf sinB = \frac{1}{2}sinA }

A B = 90 {\bf \angle_A - \angle_B = 90 \implies }

s i n A = s i n ( 90 + B ) = c o s B {\bf sinA = sin(90 + B) = cosB }

s i n B = s i n ( A 90 ) = c o s A {\bf sinB = sin(A - 90) = -cosA }

s i n A = 2 c o s A t a n A = 2 {\bf \implies sinA = -2cosA \implies tanA = -2 }

s i n B = 1 2 c o s B t a n B = 1 2 {\bf sinB = \frac{1}{2}cosB \implies tanB = \frac{1}{2} }

{\bf \implies }

t a n C = t a n ( 180 ( A + B ) ) = t a n ( A + B ) = ( t a n A + t a n B 1 t a n A t a n B ) = 3 4 . {\bf tanC = tan(180 - (A + B)) = -tan(A + B) = -(\frac{tanA + tanB}{1 - tanAtanB}) = \frac{3}{4}. }

It is given that A B = 9 0 A = 9 0 + B A - B = 90^\circ \implies A = 90^\circ + B .

C = 18 0 A B = 18 0 ( 9 0 + B ) B = 9 0 2 B \begin{aligned} \implies C & = 180^\circ - A - B \\ & = 180^\circ - (90^\circ + B) - B \\ & = 90^\circ - 2 B \end{aligned}

Using sine rule, we have:

sin A 2 = sin B 1 sin A = 2 sin B sin ( 9 0 + B ) = 2 sin B tan B = 1 2 \begin{aligned} \frac{\sin A}{2} & = \frac{\sin B}{1} \\ \implies \sin A & = 2 \sin B \\ \sin (90^\circ + B) & = 2 \sin B \\ \implies \tan B & = \frac{1}{2} \end{aligned}

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 1 2 1 + 1 2 = 1 3 = 2 ( 1 3 ) 1 ( 1 3 ) 2 = 6 8 = 3 4 = 0.75 \begin{aligned} \tan C & = \tan (90^\circ + 2 B) \\ & = \tan (2(45^\circ - B)) \quad \quad \small \color{#3D99F6}{\tan (45^\circ-B) = \frac{1-\tan B}{1+\tan B} = \frac{1-\frac{1}{2}}{1+\frac{1}{2}} = \frac{1}{3}} \\ & = \frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{3}\right)^2} = \frac{6}{8} = \frac{3}{4} = \boxed{0.75} \end{aligned}

Emrul Kais
Dec 24, 2015

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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