Question

Помогите решить три задачи. ОЧЕНЬ ПРОШУ ПОМОГИТЕ!!! 1) найдите косинус угла альфа и тангенс угла альфа, если синус угла альфа=2\3 и 0 градусов< угла альфа< 90 градусов. 2) Найдите синус и косинус угла альфа, если тангенс угла альфа равен 1\2 3) Докажите, что если два прямоугольных треугольника имеют по равному катету, то отношение синусов углов, противолежащих этим катетам, обратно отношению гипотенуз, а отношение тангенсов этих углов обратно отношению неравных катетов.

Answer 1

1.
Т.к. Син угла альфа =2/3то вс=2,а ва=3
По теореме пифагора ав^2=вс^2+ас^2;ас^2=ав^2-вс^2;ас^2=9-4;ас=5в корне.
кос=ас/ва=5в корне/3
Тангенс=2/3*(5в корне)/3=2*3/3*(5в корне)=2/(5в корне)

Answer 2

I'd be happy to help you with these three problems! Let's go through each one step by step.

Problem 1:

To find the cosine and tangent of angle alpha, given that the sine of alpha is 2/3 and the angle is between 0 and 90 degrees, we can use the Pythagorean identity and the definitions of cosine and tangent.

The Pythagorean identity states that sin^2(alpha) + cos^2(alpha) = 1. Since we know that sin(alpha) = 2/3, we can substitute this value into the equation:

(2/3)^2 + cos^2(alpha) = 1

Simplifying this equation, we get:

4/9 + cos^2(alpha) = 1

Subtracting 4/9 from both sides, we have:

cos^2(alpha) = 5/9

Taking the square root of both sides, we find:

cos(alpha) = ±√(5/9)

Since the angle alpha is between 0 and 90 degrees, the cosine of alpha must be positive. Therefore, we have:

cos(alpha) = √(5/9)

To find the tangent of alpha, we can use the definition of tangent:

tan(alpha) = sin(alpha) / cos(alpha)

Substituting the given values, we get:

tan(alpha) = (2/3) / √(5/9)

Simplifying this expression, we have:

tan(alpha) = (2/3) * √(9/5)

tan(alpha) = (2/3) * (3/√5)

tan(alpha) = 2/√5

Therefore, the cosine of angle alpha is √(5/9) and the tangent of angle alpha is 2/√5.

Problem 2:

To find the sine and cosine of angle alpha, given that the tangent of alpha is 1/2, we can use the definitions of sine, cosine, and tangent.

The definition of tangent is tan(alpha) = sin(alpha) / cos(alpha). Since we know that tan(alpha) = 1/2, we can substitute this value into the equation:

1/2 = sin(alpha) / cos(alpha)

To solve for sin(alpha), we can multiply both sides of the equation by cos(alpha):

(1/2) * cos(alpha) = sin(alpha)

Similarly, to solve for cos(alpha), we can multiply both sides of the equation by cos(alpha):

(1/2) = sin(alpha) * cos(alpha)

Using the Pythagorean identity sin^2(alpha) + cos^2(alpha) = 1, we can substitute the expression for sin(alpha) from the previous equation:

(1/2) = (sin(alpha) * cos(alpha))

Simplifying this equation, we get:

1/2 = (sin(alpha) * cos(alpha))

Since we know that sin^2(alpha) + cos^2(alpha) = 1, we can substitute the expression for cos^2(alpha) from the previous equation:

1/2 = (sin(alpha) * √(1 - sin^2(alpha)))

Simplifying this equation, we have:

1/2 = (sin(alpha) * √(1 - (sin(alpha))^2))

Squaring both sides of the equation, we get:

1/4 = (sin^2(alpha) * (1 - (sin(alpha))^2))

Expanding the equation, we have:

1/4 = (sin^2(alpha) - (sin^4(alpha)))

Rearranging the terms, we get:

(sin^4(alpha)) + (sin^2(alpha)) - (1/4) = 0

This is a quadratic equation in terms of sin^2(alpha). We can solve it by factoring:

(sin^2(alpha) - 1/2)(sin^2(alpha) + 1/2) = 0

This equation has two solutions: sin^2(alpha) = 1/2 and sin^2(alpha) = -1/2. However, since the sine of an angle is always between -1 and 1, the second solution is not valid. Therefore, we have:

sin^2(alpha) = 1/2

Taking the square root of both sides, we find:

sin(alpha) = ±√(1/2)

Since the angle alpha is between 0 and 90 degrees, the sine of alpha must be positive. Therefore, we have:

sin(alpha) = √(1/2)

To find the cosine of alpha, we can substitute the value of sin(alpha) into the equation sin^2(alpha) + cos^2(alpha) = 1:

(1/2) + cos^2(alpha) = 1

Simplifying this equation, we get:

cos^2(alpha) = 1/2

Taking the square root of both sides, we find:

cos(alpha) = ±√(1/2)

Since the angle alpha is between 0 and 90 degrees, the cosine of alpha must be positive. Therefore, we have:

cos(alpha) = √(1/2)

Therefore, the sine of angle alpha is √(1/2) and the cosine of angle alpha is √(1/2).

Problem 3:

To prove that if two right triangles have equal legs, then the ratio of the sines of the angles opposite those legs is equal to the ratio of the hypotenuses, and the ratio of the tangents of those angles is equal to the ratio of the unequal legs, we can use the definitions of sine, cosine, and tangent.

Let's consider two right triangles, Triangle ABC and Triangle DEF, with equal legs AB = DE and BC = EF.

We want to prove the following ratios:

sin(angle B) / sin(angle C) = AB / BC tan(angle B) / tan(angle C) = AB / BC

Using the definitions of sine and tangent, we have:

sin(angle B) = AB / hypotenuse AB sin(angle C) = BC / hypotenuse AB tan(angle B) = AB / BC tan(angle C) = BC / AB

Substituting these values into the ratios, we get:

(AB / hypotenuse AB) / (BC / hypotenuse AB) = AB / BC (AB / BC) / (BC / AB) = AB / BC

Simplifying both sides of the equations, we have:

1 = 1

Therefore, we have proved that if two right triangles have equal legs, then the ratio of the sines of the angles opposite those legs is equal to the ratio of the hypotenuses, and the ratio of the tangents of those angles is equal to the ratio of the unequal legs.

I hope this helps! Let me know if you have any further questions.