Question 40414
  i- ABC is an acute triangle
 ii- ABC is an isosceles traingle
iii- ABC is an obtuse triangle where B is an obtuse angle
 iv- ABC is a right triangle where A is the right angle
  v- ABC is an obtuse triagnle where A is the obtuse angle
 vi- In triangle ABC, sinA < a/c and cosA > b/c. 

How many of the above statements are always false regarding triangle ABC?
<pre><font face = "consolas" color = "indigo" size = 4><b>

iii, iv, v are NEVER true.

Here's why:

Draw the figure:

       B  
      /|
     / |
  c /  |a
   /   |
  /   _|
A/___|_|C
    b

i-ABC is an acute triangle

Yes this is always true because C is a 90° angle,
and the sum of all three angles of a triangle is
180°, so the sum of A and B must equal 90°. So each
of the other angles must be less than 90°. 

ii-ABC is an isosceles traingle

This is sometimes the case when a and b are equal and
A and B are both 45°.

iii-ABC is an obtuse triangle where B is an obtuse angle

This is NEVER the case because C is a 90° angle,
and the sum of all three angles of a triangle is
180°, but an obtuse angle is greater than 90°, so
if B were an obtuse angle, that would make the 
sum more than 180°.

iv-ABC is a right triangle where A is the right angle

This is NEVER the case because C is a 90° angle,
and the sum of all three angles of a triangle is
180°, and if A were 90°, that would make the sum
of A and C 180°, leaving 0° for B to equal, but
no triangle can have a 0° angle.

v-ABC is an obtuse triagnle where A is the obtuse angle

This is NEVER the case because C is a 90° angle,
and the sum of all three angles of a triangle is
180°, but an obtuse angle is greater than 90°, so
if A were an obtuse angle, that  would make the 
sum more than 180°.

vi- In triangle ABC, sinA < a/c and cosA > b/c.

If you use the convention where side a is opposite
angle A, side b is opposite angle B, and side c is
the hypotenuse, then this is NEVER the case because
sinA = a/c and cosA = b/c, never > or < .

However, if you vary from this notation, and have

       A  
      /|
     / |
  c /  |a
   /   |
  /   _|
B/___|_|C
    b

then sinA < a/c and cosA > a/c



Edwin
AnlytcPhil@aol.com</pre>