SOLUTION: Hi, Please help me solve this question. a is an acute angle such that sin a = {{{ 3/(2*sqrt(6)) }}} By sketching a right angled triangle, show that tan a = {{{ 3/sqrt(15) }}

Algebra ->  Trigonometry-basics -> SOLUTION: Hi, Please help me solve this question. a is an acute angle such that sin a = {{{ 3/(2*sqrt(6)) }}} By sketching a right angled triangle, show that tan a = {{{ 3/sqrt(15) }}      Log On


   



Question 1099331: Hi,
Please help me solve this question.
a is an acute angle such that sin a = +3%2F%282%2Asqrt%286%29%29+
By sketching a right angled triangle, show that tan a = +3%2Fsqrt%2815%29+
How do I do this when I don't have a side length to work out the angle with trigonometry? Please help me! I'm so confused.

Found 2 solutions by josmiceli, KMST:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If the side opposite +alpha+ is +a+ and +b+ is
the side adjacent to +alpha+ and
the hypotenuse is +c+,
+sin%28+alpha+%29+=+a%2Fc+
+a%2Fc+=+3%2F%282%2Asqrt%286%29+%29+
+b%5E2+=+c%5E2+-+a%5E2+
+b%5E2+=+%28+2%2Asqrt%286%29+%29%5E2+-+3%5E2+
+b%5E2+=+4%2A6+-+9+
+b%5E2+=+24+-+9+
+b%5E2+=+15+
+b+=+sqrt%28+15+%29+
-----------------------
+tan%28+alpha+%29+=+a%2Fb+
+tan%28+alpha+%29+=+3%2Fsqrt%2815%29+
Notice that you don't need to find the angle +alpha+

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The word "sketch" suggests that you would just draw something as an illustration,
without caring for accuracy or straightness of the lines,
maybe labeling vertices,
and based on the drawing would use your knowledge of trigonometry,
and of the Pythagorean theorem to calculate
what the values of the trigonometric ratio would be
for an ideal triangle where sin%28A%29=3%2F%282%2Asqrt%286%29%29+ exactly.
It could be something like this:
The angles of that drawing are not such that sin%28A%29=3%2F%282%2Asqrt%286%29%29+ ,
but we say it is just a "sketch" representing the problem.
Then you could say that your triangle has BC and AB side lengths
(no need to mentions units or measure sides)
that yield sin%28A%29=3%2F%282%2Asqrt%286%29%29+,
such as AB=2%2Asqrt%286%29 and BC=3 .
With the numbers above, you could calculate AC.
The Pythagorean theorem says that AC%5E2%2BBC%5E2%2BAB%5E2 ,
so AC%5E2%2B3%5E2=%282%2Asqrt%286%29%29%5E2%29
AC%5E2%2B9=2%5E2%2A%28sqrt%286%29%29%5E2%29
AC%5E2%2B9=4%2A6
AC%5E2%2B9=24
AC%5E2=24-9
AC%5E2=15
AC=sqrt%2815%29 .
Then, tan%28A%29=BC%2FAC=3%2Fsqrt%2815%29

You could also just state that sin%28A%29=BC%2FAB=3%2F%282%2Asqrt%286%29%29 ,
and work the same calculations without specific length.
It just gets to be more cumbersome.

There is an infinite number of similar right triangles that have sin%28A%29=3%2F%282%2Asqrt%286%29%29+ .
They all look like scaled up or scale down versions of one another, and they all have the same tan%28A%29=3%2Fsqrt%2815%29+ .
Saying that the one you sketched has AB=2%2Asqrt%286%29 and BC=3
is good enough for a proof in my book.

The wording of the question sounds like you are expected to draw a right triangle ABC
such that sin%28A%29=3%2F%282%2Asqrt%286%29%29+ ,
measure its legs, and calculate the tangent of A as a ratio of leg lengths.
I do not hink that is what was meant.
Doing an exact drawing without using the fact that tan%28A%29=3%2Fsqrt%2815%29+ is a bit tricky,
but I will show you what your knowledge of geometry can do for you.
First you must decide how large to make your triangle.
There is an infinite number of similar right triangles that have sin%28A%29=3%2F%282%2Asqrt%286%29%29+ .
They all look like scaled up or scale down versions of one another, and they all have the same tan%28A%29=3%2Fsqrt%2815%29+ .
Let's say you deice that the hypotenuse of your right triangle will measure 100mm.
Draw a line, set your compass for a circle of radius AB%2F2=50mm ,
and draw an arc that crosses your line at two points, A and B, like this
Now, set the compass for an arc of radius %28100mm%29%2A%283%2F%282%2Asqrt%286%29%29%29=61.2mm ,
and draw a small arc centered at B that crosses the arc you made before.
The point where the arcs cross is point C, with BC=AB%2Asin%28A%29 .

Connect point C to points A and B to get your right triangle.

The reason this works is because
1) an inscribed angle that intercepts a semicircle is always a right angle, and
2)we made AB and BC such that BC%2FAB was the value we needed for sin%28A%29 .