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
134) Side of triangle
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- sanju09
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This problem can never be seen on GMAT for the trigonometric approaches be the only way out to solve this.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
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]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
- sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
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...
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...
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com