Help me figure out area of triangle!!

**carlasader** · May 26th 2007, 09:02 AM

hey everyone , i would like you please to help me figure out 2 problems they both about figuring out the area of each a triangle ABC and an engineer estimated area of vertical cross section of water flowing under bridge. and there is a model for it .

regards

**janvdl** · May 26th 2007, 09:31 AM

Originally Posted by carlasader

hey everyone , i would like you please to help me figure out 2 problems they both about figuring out the area of each a triangle ABC and an engineer estimated area of vertical cross section of water flowing under bridge. and there is a model for it .

regards

6.a) $\frac{Sin B}{b} = \frac{Sin C}{c}$

$\frac{c Sin B}{b} = Sin C$

$C = 52,47 \ degrees$

6.b) Now angle A = 180 - (B + C)
Thus A = 52,53 degrees

Im going to work out the area now. Im just struggling a bit with it.

**janvdl** · May 26th 2007, 09:45 AM

6.c) Yes I got it!!!

Draw a line AM from A to BC, but perpendicular to BC.
So that B = 75 degrees ; A = 15 degrees ; M1 = 90 degrees

Then we can say that $cos \ 75 = \frac{BM}{AB}$

And then BM is 2,02

With Pythagoras we determine that AM(the height) = 7,534

With pythagoras again, we determine the full length of BC(the base) to be: 7,81

$Area = \frac{1}{2} (7,81) (7,534) = 29,408 \ square \ units$

**janvdl** · May 26th 2007, 10:09 AM

May i please have a picture of how the bridge looks? Im bad at visualising things.

**Soroban** · May 26th 2007, 01:37 PM

Hello, carlasader and Jhevon!

Are you familiar with this formula for the area of a triangle?

. . $Area \:=\:\frac{1}{2}\ bc\sin A$

The area is one-half the product of two sides times the sine of the included angle.

We have: . $b = AC = 9.5,\:c = AB = 7.8,\:\angle A = 52.5^o$

Hence: . $Area \:=\:\frac{1}{2}(9.5)(7.8)\sin52.5^o \:\approx\:29.4$

**earboth** · May 26th 2007, 11:47 PM

Originally Posted by carlasader

hey everyone , i would like you please to help me figure out 2 problems they both about figuring out the area of each a triangle ABC and an engineer estimated area of vertical cross section of water flowing under bridge. and there is a model for it .

regards

hello,

to #7:

if a is the lower bound of the area,
if b is the upper bound of the area,
if n is the number of trapezoids
then the trapezoid formula to estimate the value of an area is:

$A \approx \frac{b-a}{2n}(y_a + 2y_1 + 2y_3 + \ldots +2y_{n-1} + y_b)$

With your values you get:

$A \approx \frac{24-0}{2 \cdot 6}(1.2 + 2 \cdot 2.3 + 2 \cdot 3.8 + \ldots +2 \cdot 1.9 + 0.6)= 68 \ m^2$

**janvdl** · May 27th 2007, 12:03 AM

Thank you to Soroban, I completely forgot that Area formula, and once again, took a long way around.

And thank you to Earboth, i also understand that problem better now. A little bit anyway

**Jhevon** · May 27th 2007, 08:36 AM

Originally Posted by Soroban

Hello, carlasader and Jhevon!

lol, I wasn't even here Soroban

I think you meant to say "carlasader and janvdl"

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