right triangle trig

**gabriel** · August 1st 2010, 04:32 PM

ok so in this problem..
i am told to solve for all sides and angles in the right triangle
so the bottom side of the triangle is 6.2 in length
and all the other sides are unknown
in angles the the triangle is a right triangle and that angle is located on the left bottom side of the triangle, on the lower right angle the angle is 25 degrees

what are the missing sides and lengths?

**pickslides** · August 1st 2010, 05:26 PM

Find your vague description solve the missing lengths $\cos 25^{\circ} = \frac{6.2}{x}$ and $\sin 25^{\circ} = \frac{y}{6.2}$

**gabriel** · August 1st 2010, 06:10 PM

x=6.81
y=14.66
?
or y= 0.06

**skeeter** · August 2nd 2010, 05:52 AM

Originally Posted by gabriel

ok so in this problem..
i am told to solve for all sides and angles in the right triangle
so the bottom side of the triangle is 6.2 in length
and all the other sides are unknown
in angles the the triangle is a right triangle and that angle is located on the left bottom side of the triangle, on the lower right angle the angle is 25 degrees

what are the missing sides and lengths?

make a sketch, then set up a couple of trig ratios ...

$\displaystyle \tan(25^\circ) = \frac{a}{6.2}$

$\displaystyle \cos(25^\circ) = \frac{6.2}{b}$

solve for $a$ and $b$

**gabriel** · August 2nd 2010, 09:41 AM

when i solve for "a" I divide tan(25) by 6.2 and i get approximately 0.0752
when i solve for "b" I multiply cos(25) by 6.2 and i get approximately 5.62

**Archie Meade** · August 2nd 2010, 10:08 AM

Originally Posted by gabriel

when i solve for "a" I divide tan(25) by 6.2 and i get approximately 0.0752
when i solve for "b" I multiply cos(25) by 6.2 and i get approximately 5.62

From skeeter's diagram

$tan25^o=\displaystyle\huge\frac{a}{6.2}$

If both sides are equal, then 6.2 times both sides are also equal,
just as in the following..

$\frac{6}{2}=3\ \Rightarrow\ 2\frac{6}{2}=2(3)\ \Rightarrow\ \frac{2}{2}6=6$

Therefore

$(6.2)tan25^o=6.2\frac{a}{6.2}=\frac{6.2}{6.2}a=(1) a=a$

$cos25^o=\displaystyle\huge\frac{6.2}{b}\ \Rightarrow\ \frac{cos25^o}{1}=\frac{6.2}{b}$

Invert that..

$\Rightarrow\ \displaystyle\huge\frac{1}{cos25^o}=\frac{b}{6.2}$

Now multiply both sides by 6.2 to be able to calculate b.

**gabriel** · August 2nd 2010, 10:24 AM

cos(25)*6.2 = approximately 5.62
b=5.62
and (6.2)tan25 =2.89

**Archie Meade** · August 2nd 2010, 10:33 AM

Originally Posted by gabriel

cos(25)*6.2 = approximately 5.62
b=5.62
and (6.2)tan25 =2.89

Have a look at skeeter's diagram again..

is b>6.2 or is b<6.2 ?

What happens to $\frac{1}{cos25^o}=\frac{b}{6.2}$

when you multiply both sides by 6.2?

$6.2\left(\frac{1}{cos25^o}\right)=6.2\left(\frac{b }{6.2}\right)$

What is the next line?

**gabriel** · August 2nd 2010, 10:41 AM

6.2(1/cos(25) = 6.84 = b
and 6.2(6.84/6.82)=6.84

**Archie Meade** · August 2nd 2010, 11:03 AM

Originally Posted by gabriel

6.2(1/cos(25)) = 6.84 = b
and 6.2(6.84/6.2)=6.84

Yes,
but the important thing is for you to be able to think it through...

$cos25^o=\frac{6.2}{b}$

We want to find out what b is, using $cos25^o$ and 6.2.

We need b on one side and the other two combined correctly to calculate b.
There are lots of ways to combine two numbers..add, subtract, multiply, divide, etc.
but there is only one way to combine those numbers to get b.

b being in the denominator position is inconvenient,
so first multiply both sides by b.

$cos25^o=\frac{6.2}{b}\ \Rightarrow\ bcos25^o=b\left(\frac{6.2}{b}\right)=\frac{b6.2}{b }=\frac{b}{b}6.2=(1)6.2=6.2$

That gets b "above the line.
However, it's now multiplied by $cos25^o$ ,

therefore if we divide both sides by $cos25^o$ , we are left with b all alone on the left,
which is what we wanted...

$bcos25^o=6.2\ \Rightarrow\ \frac{bcos25^o}{cos25^o}=\frac{6.2}{cos25^o}\ \Rightarrow\ b(1)=b=\frac{6.2}{cos25^o}$

The important thing is to understand why you take the steps,
not just "do this", then "do that"...

Or understand the way shown earlier by inverting the fractions.

**gabriel** · August 2nd 2010, 11:13 AM

i understand know but is it the same when you involve in tangents such as side "a" where u have to apply tan(25) to find side a?

**Archie Meade** · August 2nd 2010, 11:25 AM

Originally Posted by gabriel

i understand know but is it the same when you involve in tangents such as side "a" where u have to apply tan(25) to find side a?

It all depends on where the side you are looking for is in the fraction,
so you may have to do a slightly different calculation,
but the idea is to get the side you want "by itself".

In this case....

$tan25^o=\frac{side\ opposite\ 25^o}{side\ adjacent\ 25^o}=\frac{a}{6.2}$

This time we can get "a" by itself in one step by multiplying both sides by 6.2
So it's faster!

Finding "b" takes a little longer, because we begin with it under the line.

**gabriel** · August 2nd 2010, 11:36 AM

so side "a" would be = to 2.89? if tan(25)*6.2=2.89?
the fraction should be a/6.2?

**Archie Meade** · August 2nd 2010, 11:43 AM

Originally Posted by gabriel

so side "a" would be = to 2.89? if tan(25)*6.2=2.89?
the fraction should be a/6.2?

Yes,

beginning with the fraction $tan25^o=\frac{a}{6.2}$

we only need to multiply both sides by 6.2 to get $(6.2)tan25^o=a$

Last step is to simply multiply whatever the calculator gives you for tan25 by 6.2.
That's how you figure out "a" without having to draw the triangle and measure lengths.

**gabriel** · August 2nd 2010, 11:50 AM

ok just checking

and one more question if i could
if u look back @ skeeter.s drawing, how do i find the missing angle @ the top?
would it be 90+25=115 and 180-115=65 degrees?

Thread: right triangle trig

LinkBack

Thread Tools

Display

right triangle trig

Similar Math Help Forum Discussions

Right triangle in trig

Working with Trig identies to show that a triangle is a right triangle

triangle trig

Right Triangle Trig. ?

triangle trig

Search Tags