Posted by Jasmin on Thursday, June 3, 2010 at 10:08pm.
If m<A = 45, AB = 10, and BC = 8, the greatest number of distinct triangles that can be constructed is:
11th grade - MathMate, Thursday, June 3, 2010 at 10:40pm
Angle A=45, AB=10, and BC=8.
This is an SSA configuration which could result in 0, 1 or two distinct triangles, depending on the length of the "dangling" leg BC, i.e. the side which does not touch the given angle A.
To determine which case applies, we construct a triangle where the side BC' cuts AC' at C' and make a right angle with AC'. Calculate the length BC'.
In this case angle C=90°, therefore angle B=45°. From an isosceles triangle with AB as the hypotenuse, BC'=5√2=7.071
Since BC=8 is greater than BC', there are two distinct triangles where the two possible positions of C are on each side of C'.
11th grade - Foggy, Wednesday, February 26, 2014 at 8:53pm
Answer This Question
More Related Questions
- Math - Find he number of distinct triangles with integer sides and perimeter 10 ...
- 11th grade - I'm doing The Ambiguos case of triangles given SSA. My givens are a...
- 11th grade - The students of litchfield High school are in grades 9,10,11,12. Of...
- 11th grade - what are some greatest weeknesses of caesar rodney?
- math - Are trigonometric functions taught in 11th grade, if so ,is it wrong that...
- 11th grade Algebra - A number is multiplied by -3 and then this product is ...
- geometry - Given five segments of length 2,3,5,8, and 13, what is the number of ...
- 11th grade physics - A parallel condenser of 0.8F is to be constructed using ...
- 11th grade-Calculas - The number of hours of daylight D depends upon the ...
- Language - In which sentence does the word distinct make sense? The assembly ...