In triangle RST, shown above, ∠RTS = 30° and ∠RST = 15°. What is the l

Solution:

Although triangle RST looks like a right triangle, we know it is not because of the given measures of the angles. So let’s redraw the triangle that depicts more of the actual shape of the triangle.



As we can see, triangle RST is an obtuse triangle with the obtuse angle at R. Now extend RT from R to U such that SU is perpendicular to RU. Now we’ve created two right triangles: SUT and SUR. We see that triangle SUT is a 30-60-90 right triangle and triangle SUR is a 45-45-90 right triangle.

Let’s look at triangle SUR first. Since SR, the hypotenuse of triangle SUR, is 6√2, we see that, SU and RU, the legs of triangle SUR must be each 6√2/√2 = 6. Now, let’s look at triangle SUT. Since SU, the shortest leg of triangle SUT, is 6, ST, the hypotenuse of triangle SUT, must be 6 x 2 = 12 (notice that UT = 6 + 6(√3 - 1) = 6 + 6√3 - 6 = 6√3, which is exactly √3 times SU).

Answer: B
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IMO B

This means side ST will be the greatest side among all.

SR = 6\(\sqrt{2}\) = 6*1.414 ~ 9

RT = 6(\(\sqrt{3}\) - 1) ~ 4

No ideally it can be 12, 9+4 > 12 & 5 < 12
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The value should be between the sum of the two sides and their difference. So in between 6 and 19. Only 12 fits in this category. Hence option B

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\(6\sqrt[]{2} = 6*1.4 = 8.4\)
\(6 (\sqrt[]{3}-1) = 6*(1.7-1) = 6*0.7 = 4.2\)

so the range is between 12.6 and 4.2

how come 6 and 19?

by extending RT to form the right angle triangle TZS,

ZSR is a 45:45:90 triangle with side ratios of \(1:1:\sqrt{2}\)
so SZ = 6

and SZT is a 30:60:90 triangle with side ratios of \(1:\sqrt{3}:2\)
so ST = 12

B

My bad. But we can eliminate 6 as the side opposite to the largest angle is the longest.

Actually answer is B, we can extend side RT to form a right triangle having angles ,90 45 45.

You may want to check out this post.
https://gmatclub.com/forum/in-triangle- ... l#p2223923
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(6√2)^2 + 6(√3-1)^2 = x^2
X^2 = 36*2 + 36*2
X^2 = 144
X = 12

Answer is : B

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Is that ok? I mean the method i follow?

Posted from my mobile device

I did it using a different method
we know that area of a triangle is 1/2absinΘ where Θ is the angle between the sides. Let x be the unknown side
first we find out angle SRT , it equals 135
Next we find the area using the two angles 135 and 30
considering SR and RT
angle between them is 135
so area is 1/2* 6√2*6(√3-1)* sin 135 ........(1)


Consdierinf ST and RT
angle between them is 30
so area is 1/2*x* 6(√3-1)..........(2)

equating 1 and 2 we get x = 12