*A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it.Once you've got a helpful diagram, the math is usually pretty straightforward.*

using namespace std; long long unsigned count Nums(short,short,short array[][15],short,bool,bool); int main(int argc,char **argv) long long unsigned count Nums(short start_x,short start_y,short array[][15],short size, bool goright,bool goright2) { long long unsigned current Sum; current Sum = array[start_x][start_y]; if (goright) else //go down if ((start_x 1) And it does work (I also changed the function a little so it would print every number it was going through) But I still don't get the right result.

This goes down and to the right and still goes right (adjacent numbers to the right), goes down and keeps going down (adjacent number to the left). Alastair: It's simple long long unsigned count Nums(short start_x,short start_y,short array[][15],short size, bool goright,bool goright2); start_x and start_y are the coords of the array array is the reference to the array size is just the size of the array (it's always 15) goright is to know if I am going to go down and right or just down goright2 is to know if I am going to continue to go down or going left Ok so first off, I'm a little unclear as to what you think the problem is. Secondly you might want to re-think your design here.

Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting.

For instance, if they hadn't told me to round in the exercise above, my value for the height should have been the value with the radical. That width will be twice the base of one of the right triangles.

There are two major approaches, regarding the strategy of the solution.

First, think whether the direct calculations may resolve the problem.

With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio: Standardized Test Prep ACCUPLACER Math ACT Math ASVAB Math CBEST Math CHSPE Math CLEP Math COMPASS Math FTCE Math GED Math GMAT Math GRE Math MTEL Math NES Math PERT Math PRAXIS Math SAT Math TABE Math TEAS Math TSI Math more tests...

In this lesson you will find the solutions of the typical problems that relate to the triangle sides measures.

Simplify this equation step by step and solve it: 2x 2(x - 6) (x (x-6)) = 60 (after multiplying both sides of the previous equation by 2), 2x 2x x x - 12 - 6 = 60, 6x - 18 = 60, 6x = 60 18, 6x = 78, x = 13. Hence, the second side is of 13 cm - 6 cm = 7 cm long, the third side is (13 7)/2 = 20/2 = 10 cm long in accordance with the problem condition.

You can check that the sum of the side measures is equal to the given value of the perimeter: 13 cm 7 cm 10 cm = 30 cm. The triangle side measures are 13 cm, 7 cm and 10 cm.

## Comments Triangle Problem Solving