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Given sin=3\5andcos=4\5 find tan.

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joshua1979 | Student, College Freshman | eNotes Newbie

Posted August 17, 2009 at 8:23 AM via web

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Given sin=3\5andcos=4\5 find tan.

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neela | High School Teacher | Valedictorian

Posted August 17, 2009 at 9:16 AM (Answer #1)

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Kindly allow us to rewrite the question:

(i)"Given sin35 and cos45, find tan35 and tan45".

Or

(ii)Given sin A =3/5 and cosA =4/5 find tan A.

Let us start with the basics which explains why we corrrected as above:

Definition: Consider a right angled triangle ABC with right angle at B. Then,

sineA = opposite side/hypotenuse=BC/AC,

cosine A=adjascent side/hypotenuse=AB/AC and

tangent A = opoosite side/adjascent side=BC/AB.

Note: the sineA can take values 0  to 1  when A takes values from 0 to 90 deg , sinA from  1 to 0 for A =90 deg to 180 deg, sinA from 0 to -1 for A=180 to 270 deg and  -1 to 0 for A=270 to 360 degree. Similarly the cosine and tangent ratios of an angle take values are as below:

A           =    0 to    90, 90 to   180 , 180 to  270, 270 to 360

cosA      =    1 to     0 ,   0 to   -1,   -1   to     0,   0   to     1

Tangent A= 0+infinite,  -infinte to 0,   0 to infinite,  -infinite to 0.

From the above we see that -1<=sin of an angle<=+1 and -1<=cos of an angle<=+1 and -infinite <=tan of an angle<= infinite. The sine and cosine curves are continuous, where as the graph of tangent of an angle is discontinous at 90 deg and 270 degree with infinite gap.

Therefore, the sine of an angle  cannot be 35 . Similarly, the   cosine of an angle cannot be 45. Therefore, the given question  to be meaningful should read  like :

Given sin35  and cos45,  find,   tan 35 and tan 45.

and we proceed for on the method to reach the solution:

We, know by trigonometry: (sin X)^2+(cos X)^= 1, is an identity and since (3/5)^2+(4/5)^2=1, We can say: "Given sinA=3/5 and cosA =4/5, find tan A". This is solved in second part below:

(i)

Therefore, tanx = sinX/cosX= sinX/sqrt[1-(sin X)^2]strong> or

tanX=sqrt[1-(cosX)^2]/cosX.

Let sin35 = x,

Tangent 35 = sin35/cos35 =sin35/sqrt[1-(sin35)^2]= x/(1-x^2). = 0.573576436sqrt/(1-0.573576436^2) =0.700207538

Given cos45=y,say.

Then, tangent 45 = sin45/cos45

=y/y=1.

As y=sin45 =cos45 = 1/sqrt2

(ii)

Given sinA =3/5 and cosA =4/5 , find tanA.

By trigonometry,

tanA = sinA/cosA. Substitute the given values of sinA and cosA to get the value of tan A.

tan A = (3/5)/(4/5) =3*5/(4*5) =3/4 =0.75

Hope this helps.

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texansue | High School Teacher | eNotes Newbie

Posted August 20, 2009 at 8:47 PM (Answer #2)

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The equation sin = 35 has no meaning, since the sine represents a particular ratio that is dependent on the value of the angle used.  It is impossible for the sine ratio to be larger than 1 so your equation could not have been sin x = 35.

Likewise, it is impossible to find a tangent ratio using sin 35= x and cos 45 = x because it is not possible for a unique right triangle to have two angles of 35 degrees and 45 degrees.

So, taking all that into account and since 45 + 45 = 90, I am re-writing your question so that it is possible to work the problem:

Given sin 45= x and cos 45 = x, find tan 45.

In a right triangle the sine of an angle is the ratio of the opposite leg (from the angle) to the hypotenuse.  The cosine ratio is the adjacent leg to the hypotenuse.  The tangent ratio is the opposite leg to the adjacent leg.  If one of the acute angles in a right triangle is 45 degrees, then both acute angles must be 45 degrees (so that the angle sum is 180 degrees in the triangle).  Since both acute angles are 45 degrees, then the lengths of the legs are equal (by the converse of the isosceles triangle theorem).  If  the two legs are equal length, then the ratio of the legs = 1, therefore tan 45 = 1.

 

 

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jeyaram | Student, Undergraduate | Valedictorian

Posted September 14, 2009 at 6:21 PM (Answer #3)

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i think your question was wrong, because 1>= sinA>=(-1) and 1>= cosA >=(-1) ok. if your question is sinA =3/5 & cosA =4/5 tanA=sinA/cosA =(3/5)/(4/5) =3/4 =0.75
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lochana2500 | Student, Undergraduate | Valedictorian

Posted November 11, 2012 at 6:11 PM (Answer #5)

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sinΘ = 3/5 ----(1)

cosΘ = 4/5 ----(2)

(1)/(2)

sinΘ/cosΘ = tanΘ = 3/4

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