Mixing
In $triangle ABC,$ if $2cos B = frac{a}{2}$, determine what type of triangle it is.
$begingroup$
The given options are
(a) right-angled
(b) equilateral
(c) isosceles
So, I tried using the sine rule, the cosine rule/projection formula, and even the Napier analogy, but I couldn't arrive at a proper answer. I'm guessing it's probably none of the above, but that's unfortunately not one of the given options.
trigonometry triangles
edited Feb 2 at 13:37
David K
55.7k345121
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
$endgroup$
add a comment |
$begingroup$
The given options are
(a) right-angled
(b) equilateral
(c) isosceles
So, I tried using the sine rule, the cosine rule/projection formula, and even the Napier analogy, but I couldn't arrive at a proper answer. I'm guessing it's probably none of the above, but that's unfortunately not one of the given options.
trigonometry triangles
edited Feb 2 at 13:37
David K
55.7k345121
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
$endgroup$
1
$begingroup$
Are you sure about the statement of the problem? It states that a number is equal to a length. Doesn't that seem odd to you?
$endgroup$
– José Carlos Santos
Feb 2 at 13:32
$begingroup$
If you take a right-angled triangle with hypotenuse equal to 4, then cosB =a/4. (that is possible if a<4)
$endgroup$
– asdf
Feb 2 at 13:39
add a comment |
$begingroup$
The given options are
(a) right-angled
(b) equilateral
(c) isosceles
So, I tried using the sine rule, the cosine rule/projection formula, and even the Napier analogy, but I couldn't arrive at a proper answer. I'm guessing it's probably none of the above, but that's unfortunately not one of the given options.
trigonometry triangles
edited Feb 2 at 13:37
David K
55.7k345121
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
$endgroup$
The given options are
(a) right-angled
(b) equilateral
(c) isosceles
So, I tried using the sine rule, the cosine rule/projection formula, and even the Napier analogy, but I couldn't arrive at a proper answer. I'm guessing it's probably none of the above, but that's unfortunately not one of the given options.
trigonometry triangles
trigonometry triangles
edited Feb 2 at 13:37
David K
55.7k345121
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
edited Feb 2 at 13:37
David K
55.7k345121
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
edited Feb 2 at 13:37
David K
55.7k345121
edited Feb 2 at 13:37
David K
55.7k345121
edited Feb 2 at 13:37
David K
55.7k345121
55.7k345121
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
asked Feb 2 at 13:27
Sashank SriramSashank Sriram
1444
1444
1
$begingroup$
Are you sure about the statement of the problem? It states that a number is equal to a length. Doesn't that seem odd to you?
$endgroup$
– José Carlos Santos
Feb 2 at 13:32
$begingroup$
If you take a right-angled triangle with hypotenuse equal to 4, then cosB =a/4. (that is possible if a<4)
$endgroup$
– asdf
Feb 2 at 13:39
add a comment |
1
$begingroup$
Are you sure about the statement of the problem? It states that a number is equal to a length. Doesn't that seem odd to you?
$endgroup$
– José Carlos Santos
Feb 2 at 13:32
$begingroup$
If you take a right-angled triangle with hypotenuse equal to 4, then cosB =a/4. (that is possible if a<4)
$endgroup$
– asdf
Feb 2 at 13:39
1
1
$begingroup$
Are you sure about the statement of the problem? It states that a number is equal to a length. Doesn't that seem odd to you?
$endgroup$
– José Carlos Santos
Feb 2 at 13:32
$begingroup$
Are you sure about the statement of the problem? It states that a number is equal to a length. Doesn't that seem odd to you?
$endgroup$
– José Carlos Santos
Feb 2 at 13:32
$begingroup$
If you take a right-angled triangle with hypotenuse equal to 4, then cosB =a/4. (that is possible if a<4)
$endgroup$
– asdf
Feb 2 at 13:39
$begingroup$
If you take a right-angled triangle with hypotenuse equal to 4, then cosB =a/4. (that is possible if a<4)
$endgroup$
– asdf
Feb 2 at 13:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An right angled triangle with hypotenuse equal to 4 units will satisfy the above equation.
Thus we can have an right angled triangle or an isoceles right angled triangle.
So the correct answers are A or A&C
answered Feb 2 at 13:59
Cute BabyCute Baby
11
$endgroup$
add a comment |
$begingroup$
This gives us cos(B)=a/4 . Using cos rule we get
a^2 +c^2 - b^2 / 2ac = a/4
Solving the quadratic for real value of c we get
a^2(a^2 - 8)≥ 8b^2
We see that this gives a valid value for a when a=B
It can also be right angled with hypotenuse 4
So the answer is A and C
also can be equilateral when a=2
answered Feb 2 at 14:11
BatmanBatman
12
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An right angled triangle with hypotenuse equal to 4 units will satisfy the above equation.
Thus we can have an right angled triangle or an isoceles right angled triangle.
So the correct answers are A or A&C
answered Feb 2 at 13:59
Cute BabyCute Baby
11
$endgroup$
add a comment |
$begingroup$
An right angled triangle with hypotenuse equal to 4 units will satisfy the above equation.
Thus we can have an right angled triangle or an isoceles right angled triangle.
So the correct answers are A or A&C
answered Feb 2 at 13:59
Cute BabyCute Baby
11
$endgroup$
add a comment |
$begingroup$
An right angled triangle with hypotenuse equal to 4 units will satisfy the above equation.
Thus we can have an right angled triangle or an isoceles right angled triangle.
So the correct answers are A or A&C
answered Feb 2 at 13:59
Cute BabyCute Baby
11
$endgroup$
An right angled triangle with hypotenuse equal to 4 units will satisfy the above equation.
Thus we can have an right angled triangle or an isoceles right angled triangle.
So the correct answers are A or A&C
answered Feb 2 at 13:59
Cute BabyCute Baby
11
answered Feb 2 at 13:59
Cute BabyCute Baby
11
answered Feb 2 at 13:59
Cute BabyCute Baby
11
answered Feb 2 at 13:59
Cute BabyCute Baby
11
11
add a comment |
add a comment |
$begingroup$
This gives us cos(B)=a/4 . Using cos rule we get
a^2 +c^2 - b^2 / 2ac = a/4
Solving the quadratic for real value of c we get
a^2(a^2 - 8)≥ 8b^2
We see that this gives a valid value for a when a=B
It can also be right angled with hypotenuse 4
So the answer is A and C
also can be equilateral when a=2
answered Feb 2 at 14:11
BatmanBatman
12
$endgroup$
add a comment |
$begingroup$
This gives us cos(B)=a/4 . Using cos rule we get
a^2 +c^2 - b^2 / 2ac = a/4
Solving the quadratic for real value of c we get
a^2(a^2 - 8)≥ 8b^2
We see that this gives a valid value for a when a=B
It can also be right angled with hypotenuse 4
So the answer is A and C
also can be equilateral when a=2
answered Feb 2 at 14:11
BatmanBatman
12
$endgroup$
add a comment |
$begingroup$
This gives us cos(B)=a/4 . Using cos rule we get
a^2 +c^2 - b^2 / 2ac = a/4
Solving the quadratic for real value of c we get
a^2(a^2 - 8)≥ 8b^2
We see that this gives a valid value for a when a=B
It can also be right angled with hypotenuse 4
So the answer is A and C
also can be equilateral when a=2
answered Feb 2 at 14:11
BatmanBatman
12
$endgroup$
This gives us cos(B)=a/4 . Using cos rule we get
a^2 +c^2 - b^2 / 2ac = a/4
Solving the quadratic for real value of c we get
a^2(a^2 - 8)≥ 8b^2
We see that this gives a valid value for a when a=B
It can also be right angled with hypotenuse 4
So the answer is A and C
also can be equilateral when a=2
answered Feb 2 at 14:11
BatmanBatman
12
edited Feb 3 at 8:33
answered Feb 2 at 14:11
BatmanBatman
12
answered Feb 2 at 14:11
BatmanBatman
12
answered Feb 2 at 14:11
BatmanBatman
12
12
add a comment |
add a comment |
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1
$begingroup$
Are you sure about the statement of the problem? It states that a number is equal to a length. Doesn't that seem odd to you?
$endgroup$
– José Carlos Santos
Feb 2 at 13:32
$begingroup$
If you take a right-angled triangle with hypotenuse equal to 4, then cosB =a/4. (that is possible if a<4)
$endgroup$
– asdf
Feb 2 at 13:39