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Does a right triangle exist where sin A = sin B = cos A = cos B, where A and B are the two acute angles of the triangle? Provide evidence to support your conclusion, and if possible, find the measures of angles A and B.

User Yan
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1 Answer

5 votes

Answer:

Explanation:

When you list the trigonometic ratios for both acute angles in a right triangle, you notice some interesting developments.

Let's concentrate on sine and cosine:

Observation from the table above: notice1

In a right triangle, the sine of one acute angle, A, equals the cosine of the other acute angle, B.

If ∠A and ∠B are the acute angles of a right triangle,

sin A = cos B

Since the measures of these acute angles of a right triangle add to 90º, we know these acute angles are complementary. ∠A is the complement of ∠B, and ∠B is the complement of ∠A.

If we write, m∠B = 90º - m∠A (or m∠A = 90º - m∠B ), and we substitute into the original observation, we have:

notice4

If θ is substituted for m∠A , we establish the following relationship between sine and cosine:

sincosrule

The sine of any acute angle is equal to the cosine of its complement.

The cosine of any acute angle is equal to the sine of its complement.

Sine and cosine are called "cofunctions", where the sine (or cosine) function

of any acute angle equals its cofunction of the angle's complement.

FYI: What about tangent?

Yes, there is a "relationship" regarding the tangent of the two acute angles (A and B) in a right triangle. It is not, however, the same type of relationship that exists between sine and cosine.

The tangent of ∠A is the reciprocal (flip over) of the tangent of ∠B.

The angles being used in the following examples are acute angles.

ex1

If sin 30º = ½ and cos θ = ½,

find θ. Solution: Since both trig functions = ½,

30º and θ must be complementary.

θ = 60º

ex2

If sin(3x + 10)º = cos(x + 24)º,

find x. Solution: In order for the sine and cosine to be equal, the angles must be complementary.

3x + 10 + x + 24 = 90

4x + 34 = 90

4x = 56

x = 14

ex3

If sin(15º) = 0.26 and

cos (15º) = 0.97,

find sin(75º) and the cos(75º). Solution: The sine of an angle and the cosine of its complement are equal.

15º and 75º are complementary.

sin(75º) = cos(15º) = 0.97

cos(75º) = sin(15º) = 0.26

ex4

In right ΔABC, m∠C = 90º,

sin A = x + 0.1 and cos B = 2x - 0.4. Find x. Solution: If A and B are the acute angles of a right triangle, sin A = cos B.

x + 0.1 = 2x - 0.4

0.5 = x

User Ganapathy C
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