# How to solve this triangle?

Please observe the triangle shown in the figure below (In red). I consider the angle between $5$ and $4$. The base should be $5$, hypotenuse $4$ and perpendicular $3$, but according to the solution manual the base is $4$ and the hypotenuse $5$. Please tell me how? Thanks! • there is no picture – anonymous Nov 13 '10 at 18:41
• Sorry,forgot to upload the picture.My bad. – user2857 Nov 13 '10 at 18:44
• The hypotenuse is the longest side of a right triangle, the side opposite the right angle. In your picture, it looks like a right angle symbol is used for an acute-looking angle that is also adjacent to the longest side. Where did the picture come from? – Jonas Meyer Nov 13 '10 at 18:47
• I edited the picture,added the angle that i considered – user2857 Nov 13 '10 at 18:54
• It isn't clear from the picture that it is a right triangle. But if it is, the hypotenuse should be the longest side and opposite the right angle. (More specifically, the square of the length of the hypotenuse is the sum of the squares of the other two sides.) – Jonas Meyer Nov 13 '10 at 19:03

This is not a complete answer to your question, but here's a hint. (I cannot post comments yet). (3,5,4) is not a Pythagorean triplet i.e. there are no right angled triangle with base 5, perpendicular 3 and hypoteneuse 4.

By definition, the hypothenuse is the longest side of a right triangle; so in any case you could not possibly have a triangle in which the "base" is larger than the hypothenuse as you claim it "should" be here. In any case, "base" depends on orientation (so it is possible for the hypothenuse to be the base).

Your picture is misleading you somehow into thinking that the side opposite the right angle is the side of length $4$; it cannot be.

• My interpretation of the picture is: 1) the origine of the coordinate system $x,y,z$ is the centre of the piece "eye" 2) force $F_1$ (with a magnitude of $80$ N) lies in the $x,y$ plane 3) its direction is $(x,y,z)=(4,0,3)$ 4) the angle between the force components $F_x,F_z$ is $\pi/2$ 5) the force component $F_z=0$ 6) the side (of the right triangle) with length $5$ is opposite the right angle formed by the sides with lengths $3$ and $4$. – Américo Tavares Nov 13 '10 at 23:39
• Typo: 2) ... lies in the $x,z$ plane. – Américo Tavares Nov 14 '10 at 10:58

This is a transcrition of my comment:

My interpretation of the picture is:

1) the origine of the coordinate system $x,y,z$ is the centre of the piece "eye"

2) force $F_1$ (with a magnitude of 80 N) lies in the $x,z$ plane

3) its direction is $(x,y,z)=(4,0,3)$

4) the angle between the force components $F_{1_{x}},F_{1_{z}}$ is $π/2$

5) the force component $F_{1_{z}}=0$

6) the side (of the right triangle) with length 5 is opposite the right angle formed by the sides with lengths 3 and 4.

The triangle lies in the $x,z$ plane. Its vertices are $P(0,0,0)$, $Q(4,0,0)$ and $R(4,0,3)$. The hypotenuse is $PR$ and the base $PQ$.