Dear Sir/Madam;
Kindly help me with this problem, I am not familiar with this type of problem. here it is:
A triangular lot has two of its angles measuring 25 degrees and 79 degrees respectively. The longest side has a measure of 26 meters. What are the measures of the other two sides?
Thank you very much for your help!
Sketch ΔABC, with angle A, 25o, and angle B, 79o. Now, since angles A and B total 104o, then angle C MUST measure 76o.
Thus, angle B is the largest angle, which means that side AC, or b, which is OPPOSITE angle B is the longest side,
at 26 meters.
The Law of Sines can then be used to determine sides BC (a), and AB (c), the shortest and middle sides
Determining side BC, or a (shortest side)
a * sin 79 = 26 * sin 25 ------- Cross-multiplying
a =
a, or side BC, or shortest side = 11.1937 ≈ m
Determining side AB, or c (middle side)
c * sin 79 = 26 * sin 76 ------- Cross-multiplying
c =
c, or side AB, or middle side = 25.699 ≈ m
You can do the check!!
Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.
Further help is available, online or in-person, for a fee, obviously.
For FREE info and answers to questions about the ASVAB exam, the NYS 3 – 8 city/state wide exams,
GENERAL MATH QUESTIONS, MATH HOMEWORK QUESTIONS, or MATH QUESTIONS related to the Regents Integrated Algebra,
Regents Geometry, Regents Algebra 2/Trigonometry, SHSAT, COOP/HSPT/TACHS, PSAT, SAT, ACT, SSAT/ISEE, GRE, CLEP,
and the GED, you can visit: http://asvabstudyzone.freeforums.net/.