134) Side of triangle

ern5231 wrote:I recollect seeing this question online but don't recollect where it is now. So pardon me for putting out the question like this. How can we solve this. I am not quite able to figure it out

A triangle ABC has two angles 45 and 60. The side to the right side of 45 degree angle is 5 cm. Find the length of the angle to the right side of the 60 degree angle.
a) 5 root (2) b) 5 c) 5 root (6) d) 5 root(3) e) 6

This problem can never be seen on GMAT for the trigonometric approaches be the only way out to solve this.

The favorite sine rule and/or the cosine formula provided with the tables, slide rule, or calculator can be of great help here. If we ignore the small errors in the question, then we can have a triangle ABC in which angle A is 60 degrees, angle B is 45 degrees, and hence angle C is 75 degrees, side BC = 5, and side AB =?

Sine Rule

5/sin 60 = AB/sin 75

AB = 5 sin 75/sin 60 = 5 [(√3 + 1)/2 √2]/(√3/2)

AB = [spoiler]5 (3 + √3)/√2, which I am not sure same as any option here.[/spoiler]

Wait a while…

Let me correct myself…

Let’s have a triangle ABC in which angle A is 60 degrees, angle B is 45 degrees, and hence angle C is 75 degrees, side BC = 5, and side AB =?

Let’s now draw CD perpendicular to AB that meets AB in D.

Now have BD = CD = 5/√2 from the 45˚-45˚-90˚ triangle BCD.

Then have AD = CD/√3 = 5/√6 from the 30˚-60˚-90˚ triangle CAD.

Now, AB = AD + DB = 5/√6 + 5/√2

AB = 5√2 (√3 + 3)/6, in its most wanted rational denominator form.

Sorry! This could be a GMAT problem, no Trigonometry is really required.

My apologies…