geometry obtuse triangle

francoisph wrote:What is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between two sides is 1500?

any ideas please ?

i hope its not 1500 but 150.

Simple !!
formula is 1/2 * a* b* sinC
= 1/2 * 8 * 12 * sin150
=48 * sin30
=24

Let me know if you have any questions!!

yeah 150°

how to have the value Sin150?

francoisph wrote:yeah 150°

how to have the value Sin150?

hmm.. Thats a tough question.. we will need to know the sin and cos values of 0,30,45,90,120,150,180,270,360

These will be useful when dealing with trigonometry..

you can see this link for reference..

https://www.analyzemath.com/trigonometry ... gles_1.gif

I cant believe that GMAT test us about sin and cos

I didnt do this stuff long years ago, it is nightmare

francoisph wrote:I cant believe that GMAT test us about sin and cos

I didnt do this stuff long years ago, it is nightmare

Thats true.. But SIN and COS values for 30,45,60,90 would be sufficient enough.
I dont think we can do this problem without using any trigonometric functions..

I will post if i can find any way..

francoisph,

IMHO GMAT wont test trigonometric function.Usually geometry prob can be solved by using basic formulas.

What is the source of this question

Hey, while you don't need to remember any formula, I'm not sure either how you can solve this without a basic knowledge of trignometry.

IMO you do need to know the value of sin 30 & a basic understanding of what sin, cos and tan are for this sum.



If you look at the figure I've attached, let the red line be p. Then: sin 30 = 8/p. Therefore p=4.

Hence area will be 1/2 * b * h = 1/2 * 12*8 = 24

jube's got it!

You guys are correct: the GMAT does not test trig functions. However, this question can be solved without trig, as jube pointed out.

All you need to know is the proportions of a 30-60-90 triangle (which is DEFINITELY tested on the GMAT). The proportion is:

x : x*sqrt(3) : 2x. That's for the sides opposite the 30-, 60-, and 90-degree angles, respectively.

Notice that this means the hypotenuse is twice the length of the shortest side. So the imaginary triangle in jube's figure is a 30-60-90 triangle with a hypotenuse of 8 and a height of 4 (since the chosen height is opposite the 30-degree angle).

That height is the same height of the obtuse triangle. So for the obtuse triangle, base is 12 and height is 4, making for an area of 12*4 / 2 or 24.