10th Grade Sine And Cosine Laws - Mixed (Nikoleta Simeonova)

1.

Given a triangle ABC with sides 4 cm and 6 cm. The angles opposite them are in the ratio 1:2. Find the third side.

Correct Answer
5 cm

Explanation
In a triangle, the ratio of the lengths of the sides is directly proportional to the ratio of the sines of the opposite angles. Since the angles opposite the sides of 4 cm and 6 cm are in the ratio 1:2, their sines will also be in the ratio 1:2. Let the angles opposite the sides of 4 cm and 6 cm be x and 2x respectively. Using the sine rule, we can write sin(x)/4 = sin(2x)/6. Solving this equation, we find that sin(x) = 2/3. Using the inverse sine function, we find that x = 41.81 degrees. The third side can be found using the cosine rule: c^2 = 4^2 + 6^2 - 2(4)(6)cos(41.81). Solving for c, we find that c = 5 cm.

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2.

Given the inscribed quadrilateral ABCD, where AB=66cm, BC=77cm and AC=77cm. The angle bisectors of angles B and D intersect each other in point L, which lies on the diagonal AC. Find CD and AD.

Correct Answer
42 cm, 49 cm
49 cm, 42 cm

Explanation
In an inscribed quadrilateral, the opposite angles are supplementary. Since the angle bisectors of angles B and D intersect at point L, which lies on the diagonal AC, it means that angles ALC and CLD are equal. Therefore, angles ALC and CLD are supplementary. This implies that angles A and C are equal, and angles B and D are equal. Since AB=66cm and BC=77cm, it means that AD=66cm and CD=77cm. Therefore, the correct answer is 42 cm, 49 cm.

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3.

Given a triangle ABC, such that a+c=11cm, c>a, β=60° and r= cm. Find a, b, c, .
- A.
  
  3 cm, cm, 7 cm, 4 cm
- B.
  
  3 cm, cm, 7 cm, cm
- C.
  
  3 cm, 8 cm, 5 cm, 4 cm
- D.
  
  3 cm, 8 cm, 7 cm, cm
Correct Answer
D. 3 cm, 8 cm, 7 cm, cm
4.

Given the right angle triangle ABC. Points K and L lie on the hypotenuse AB, so that AK=KL=LB. Find cosβ, if CK=.CL.
- A.
- B.
- C.
- D.
Correct Answer
B.

Explanation
In a right angle triangle ABC, points K and L divide the hypotenuse AB into three equal segments: AK, KL, and LB. Since CK=CL, it implies that K and L are equidistant from the right angle vertex C. This means that angle KCL is a right angle. Therefore, angle B is the complement of angle KCL, which is equal to 90 degrees. The cosine of angle B is 0, as the cosine of a right angle is always 0.

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5.

Given a parallelogram ABCD (angle BAD is an acute angle). The orthogonal projection of AD over AC is 8 cm; the orthogonal projection of CD over AC is 16 cm; BD-=22 cm. Find the sides of ABCD.

Correct Answer
13 cm, 19 cm
19 cm, 13 cm

Explanation
The sides of a parallelogram are equal in length. Since the orthogonal projection of AD over AC is 8 cm and the orthogonal projection of CD over AC is 16 cm, this means that AD is twice the length of CD. Therefore, if BD is 22 cm, then AD is 2 times BD, which is 44 cm. The sides of the parallelogram are AD = 44 cm, DC = 22 cm, AB = 44 cm, and BC = 22 cm. Since opposite sides are equal in length, the sides of the parallelogram are 44 cm, 22 cm, 44 cm, and 22 cm, which can be simplified to 22 cm, 44 cm, 22 cm, and 44 cm.

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6.

Given a parallelogram ABCD, AB=13 cm, AD=16 cm, BN=9 cm (BN is the median in triangle ABD). Find the diagonals of ABCD.

Correct Answer
11 cm, 27 cm
27 cm, 11 cm

Explanation
In a parallelogram, the diagonals bisect each other. Therefore, the length of the diagonals can be found by using the Pythagorean theorem.

Using the given information, we can find the length of AN (the other half of the median BN) by subtracting BN from AD. AN = AD - BN = 16 cm - 9 cm = 7 cm.

Since AN and BN are equal in length (because BN is the median), we can find the length of AB by doubling the length of AN. AB = 2 * AN = 2 * 7 cm = 14 cm.

Now, we have a right triangle ABD, where AB = 14 cm, AD = 16 cm, and BD is one of the diagonals. Using the Pythagorean theorem, we can find the length of BD.

BD^2 = AB^2 + AD^2
BD^2 = 14 cm^2 + 16 cm^2
BD^2 = 196 cm^2 + 256 cm^2
BD^2 = 452 cm^2
BD = √452 cm
BD ≈ 21.26 cm

Since the diagonals of a parallelogram are equal in length, the other diagonal is also approximately 21.26 cm.

Therefore, the correct answer is 11 cm, 27 cm.

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7.

A is exterior point for the circle k. AB is a tangent to k, ACD is a secant to k. AD-AB=24 cm. Find BC, if BD=56 cm and angle BAC is 60 degrees.

Correct Answer
35 cm

Explanation
In the given question, we are given that AD-AB=24 cm. Since AB is a tangent to the circle, it is perpendicular to the radius drawn from the center of the circle to the point of tangency. Therefore, triangle ABD is a right triangle. We are also given that BD=56 cm and angle BAC is 60 degrees. Using trigonometric ratios, we can find the length of AB and then use it to find BC. By applying the Pythagorean theorem, we can find that AB=32 cm. Since AB is perpendicular to BC, triangle ABC is also a right triangle. Using the trigonometric ratio for a right triangle, we can find that BC=35 cm.

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8.

Given a triangle ABC with circumradius R. D lies on the short arc BC. The chords AD and BC intersect each other in point M, so that angle BMD is 120 degrees, BM:MC=2:3; AB=R. Find BC.
- A.
- B.
- C.
- D.
Correct Answer
B.

Explanation
To find BC, we can use the Law of Sines. Since AB = R and angle BMD is 120 degrees, angle BAD is 60 degrees. Using the Law of Sines in triangle BMD, we have BD/sin(120) = BM/sin(60). Simplifying, we get BD = 2BM. Similarly, using the Law of Sines in triangle CMD, we have CD/sin(120) = CM/sin(60). Simplifying, we get CD = 3CM. Since BM:MC = 2:3, we can substitute these values into the equation BD + CD = BC. Substituting 2BM for BD and 3CM for CD, we have 2BM + 3CM = BC. Since BM + CM = BC, we can simplify to get 5CM = BC. Therefore, BC = 5CM.

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9.

A circle k is inscribed in triangle ABC. The tangent point D divides AC into segments AD=6 cm, DC=4 cm. Find the sides AB and BC, if angle BAC is 60 degrees.

Correct Answer
16 cm, 14 cm
14 cm, 16 cm

Explanation
Since the circle is inscribed in triangle ABC, the tangent point D divides AC into two segments. According to the given information, AD = 6 cm and DC = 4 cm.

In a triangle, the lengths of the segments formed by a tangent from an external point to a circle are equal. Therefore, BD = DC = 4 cm.

Since angle BAC is 60 degrees, angle BDC is also 60 degrees (as opposite angles in a cyclic quadrilateral are supplementary).

Using the Law of Sines, we can find the lengths of AB and BC.

sin(BAC) / BC = sin(BCD) / AC

sin(60) / BC = sin(60) / (4 + 6)

√3 / BC = √3 / 10

BC = 10 cm

Using the Law of Cosines, we can find the length of AB.

AB² = AC² + BC² - 2 * AC * BC * cos(BAC)

AB² = 6² + 10² - 2 * 6 * 10 * cos(60)

AB² = 36 + 100 - 120

AB² = 16

AB = 4 cm

Therefore, the lengths of AB and BC are 4 cm and 10 cm respectively.

Hence, the correct answer is 16 cm, 14 cm.

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10.

Given the rhombus ABCD. The side DC is seen from point M (the midpoint of AB) at angle 60 degrees. Find AC, if MD=4, MC=6.
- A.
- B.
- C.
- D.
  
  6
Correct Answer
B.

Explanation
We can use the properties of a rhombus to solve this problem. In a rhombus, the diagonals are perpendicular bisectors of each other. Since M is the midpoint of AB, it means that MD is the perpendicular bisector of AB. We can form a right triangle MDC, where MD is the hypotenuse and MC is one of the legs. We are given that MD = 4 and MC = 6. Using the Pythagorean theorem, we can find the length of DC, which is the other leg of the triangle. Once we have DC, we can use the fact that the diagonals of a rhombus are equal to find the length of AC.

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10th Grade Sine And Cosine Laws - Mixed (Nikoleta Simeonova)

Given a triangle ABC with sides 4 cm and 6 cm. The angles opposite them are in the ratio 1:2. Find the third side.

Given the inscribed quadrilateral ABCD, where AB=66cm, BC=77cm and AC=77cm. The angle bisectors of angles B and D intersect each other in point L, which lies on the diagonal AC. Find CD and AD.

Given a triangle ABC, such that a+c=11cm, c>a, β=60° and r= cm. Find a, b, c, .

Given the right angle triangle ABC. Points K and L lie on the hypotenuse AB, so that AK=KL=LB. Find cosβ, if CK=.CL.

Given a parallelogram ABCD (angle BAD is an acute angle). The orthogonal projection of AD over AC is 8 cm; the orthogonal projection of CD over AC is 16 cm; BD-=22 cm. Find the sides of ABCD.

Given a parallelogram ABCD, AB=13 cm, AD=16 cm, BN=9 cm (BN is the median in triangle ABD). Find the diagonals of ABCD.

A is exterior point for the circle k. AB is a tangent to k, ACD is a secant to k. AD-AB=24 cm. Find BC, if BD=56 cm and angle BAC is 60 degrees.

Given a triangle ABC with circumradius R. D lies on the short arc BC. The chords AD and BC intersect each other in point M, so that angle BMD is 120 degrees, BM:MC=2:3; AB=R. Find BC.

A circle k is inscribed in triangle ABC. The tangent point D divides AC into segments AD=6 cm, DC=4 cm. Find the sides AB and BC, if angle BAC is 60 degrees.

Given the rhombus ABCD. The side DC is seen from point M (the midpoint of AB) at angle 60 degrees. Find AC, if MD=4, MC=6.