What is the area of an obtuse angled triangle whose two

Question

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?

A. 24 sq units
B. 48 sq units
C. 24V3
D. 48V3
E. Such a triangle does not exist

maratikus · Answer

(8+6sqrt(3))*6/2 - 6*6sqrt(3)/2 = 8*3 = 24 -> A

abhijit_sen · Answer

Question seems to be bit weird to me.
What does it mean by angle included betweem them in 1500.

Sum of all the angles in a triangle is supposed to be 180, then how come 1 angle can be 1500.

So I will go with E. Such a triangle does not exist.

alpha_plus_gamma · Answer

Its E. Such a triangle does not exist.

I hope 1500 isn't typo

sreehari · Answer

EDIT: I have corrected my calculation and now getting a different answer

1500 = 4*360 + 60 => 4*360 can be ignored because it's basically circling around the point.

So now we have two sides with lengths 8 and 12 with 60 degrees between them

Area of the triangle = (1/2)*8*12*Sin(60) = 24*sqrt(3)

I am getting C. Is something wrong with teh above approach?

neelesh · Answer

I think the question stem meant  150^o

sreehari · Answer

Also how can 1500->60 can be an obtuse angle as the question says? I think 150 makes sense.

In case of 150, Sin(150)=1/2 using the formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B) 
(I have been brushing up my trigonometry skills to solve this problem)
So the area becomes 24

What is OA?

Vemuri · Answer

The OA is "A". Give yourself a pat on the back if you got it right :-)

pmenon · Answer

can someone kindly explain this one out ? a picture will probably help as well

maratikus · Answer

I'm not good at drawing. I'll use coordinates. A(0,0), B(8,0), C(-12*sqrt(3)/2,12*1/2)=(-6*sqrt(3),6)

H(-6*sqrt(3),0) is going to be perpendicular to AB (though it's going to be outside AB, HC = 6), and area of ABC is equal to the area of HBC - area of HAC, both of those are right angle triangles. Area of HBC = (1/2)*HB*HC = (1/2)*(HA+AB)*HC. area of HAC = (1/2)*HC*HA

Area of HBC - Area of HAC = (1/2)*HC*AB = (1/2)*6*8 = 24

Let me know if that helps.

sreehari · Answer

maratikus - can you explain how you calculated HC? Let us use the below picture. So you used AB=8 and AC=12. My understanding from >360 angles is that you can subtract n*360 from the angle because they are circling around a point. Now that makes the angle included between two sides 60 (1500 - 4*360). So originally I asusmed BC=12, and Angle ABC = 60. So that way area came to be 24*sqrt(3). Did I miss something here? Looks like you sed 60 for angle HAC, how did you arrive at that? Btw a shorter way to calculate area is (1/2)*HC*AB without making any subtractions. Condition is the base from which the height is calculated should be used to calculate the area too. untitled.JPG

maratikus · Answer

the angle between two sides that are equal to 8 and 12 is 150 degrees, therefore in your picture BC = 12 would not work. AC = 12, AB = 8 (or AC = 8 and AB = 12) - I chose the first one

you are right about directly calculating triangle area - i haven't solved geometric problems for a while

Since BAC = 150 -> HAC = 30 -> sin(HAC) = 1/2 -> HC = AC*sin(HAC) = 6

sreehari · Answer

Thanks maratikus!

So the angle is 150, not 1500 as mentioned in the question. I think that's what is causing all the confusion.

What is the area of an obtuse angled triangle whose two

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