In a right-angled triangle ABC, angle C = 90 degrees, cos A = 0.4. Find sin A 6 In right triangle ABC

| March 2, 2021 | education

In a right-angled triangle ABC, angle C = 90 degrees, cos A = 0.4. Find sin A 6 In right triangle ABC, angle C = 90, sin A = √15 / 4, BC = √5. Find AB 10. The sides of the rectangle are 6 and 2 √3. Find the corners that form a diagonal with the sides of the rectangle.

1. Let us express the cosine of the angle A: AC / AB = 0.4. Therefore, the AC leg refers to the AB hypotenuse as 4 to 10.

By the Pythagorean theorem: BC² = AB² – AC² = 10² – 4² = 100 – 16 = 84. BC = √84 = 2√21.

That is, the leg BC refers to the hypotenuse AB as 2√21 to 10. This means that the sine of the angle A = 2√21 / 10 = √21 / 5.

2. Let us express the sine of the angle A: sinA = BC / AB = √15 / 4. Since BC = √5, the equation is obtained:

√5 / AB = √15 / 4.

Hence AB = 4√5 / √15 = 4√5 / (√3 * √5) = 4 / √3.

3. Let ABCD be a given rectangle. AB = 2√3, BC = 6. Find the angle ABD and the angle ADB.

Consider triangle ABD: angle BAD = 90 ° (ABCD – rectangle), then triangle ABD is rectangular.

Let us find the diagonal BD by the Pythagorean theorem:

BD² = AB² + AD² = 6² + (2√3) ² = 36 + 12 = 48. BD = √48 = 4√3.

Let us express the sine of the angle ABD: sinABD = AD / BD = 6 / 4√3 = 3 / 2√3 = 3√3 / (2 * 3) = √3 / 2. This means that the angle ABD is 60 °.

Therefore, the angle ADB = 180 ° – (90 ° + 60 °) = 30 ° (the sum of the angles in the triangle is 180 °).



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