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C2 Qs on sine/cosine rule...pls help!!!
1) The area of a triangle is 10 cm^2. The angle between two of the sides, of length 6 cm and 8 cm respectively, is obtuse. Work out:
a) The size of this angle.
b) The length of the third side.
2) The sides of a triangle are 3 cm, 5 cm and 7 cm respectively. Show that the largest angle is 120 degrees, and find the are of the triangle.
5) In triangle ABC, AB=10 cm, BC= a*sqrt 3 cm, AC= 5*sqrt 13 cm, and angle ABC=150 degrees. Calculate:
a) The value of a.
b) The exact area of triangle ABC.
6) In a triangle, the largest side has length 2 cm and one of the other sides has length sqrt 2 cm. Given that the area of the triangle is 1 cm^2, show that the triangle is right-angled and isosceles.
7) The three points, A, B and C, with coordinates A (0,1), B (3,4) and C (1,3) respectively, are joined to form a triangle:
a) Show that cos angle ACB= -4/5.
b) Calculate the are of triangle ABC.
8) The longest side of a triangle has length (2x-1) cm. The other sides have lengths (x-1) cm and (x+1) cm. Given that the largest angle is 120 degrees, work out:
a) The value of x
b) The area of the triangle.
a) The size of this angle.
b) The length of the third side.
2) The sides of a triangle are 3 cm, 5 cm and 7 cm respectively. Show that the largest angle is 120 degrees, and find the are of the triangle.
5) In triangle ABC, AB=10 cm, BC= a*sqrt 3 cm, AC= 5*sqrt 13 cm, and angle ABC=150 degrees. Calculate:
a) The value of a.
b) The exact area of triangle ABC.
6) In a triangle, the largest side has length 2 cm and one of the other sides has length sqrt 2 cm. Given that the area of the triangle is 1 cm^2, show that the triangle is right-angled and isosceles.
7) The three points, A, B and C, with coordinates A (0,1), B (3,4) and C (1,3) respectively, are joined to form a triangle:
a) Show that cos angle ACB= -4/5.
b) Calculate the are of triangle ABC.
8) The longest side of a triangle has length (2x-1) cm. The other sides have lengths (x-1) cm and (x+1) cm. Given that the largest angle is 120 degrees, work out:
a) The value of x
b) The area of the triangle.
1.(a) call the angle theta. use the sine rule for the area (area = 1/2(ab)SinC), calling C theta, and solve for theta. You know the two lengths a and b, and the area.
(b) use the cosine rule. you now know theta, and still know the sides a and b. well, c^2 = a^2 + b^2 - 2ab*cosC.
2. Use the cosine rule, and substitute in the three lengths, a, b and c, and rearrange to find theta. alternatively, rearrange algebraically and then substitute the numbers in.
5. Use the cosine rule again. Draw a diagram, label the sides a,b,c and angles(corners) A, B and C so that angle A is opposite side a. figure out how the cosine rule will help you. it has 4 variables, and you know 3 of them. use it to find the 4th (side b, or AC). then equate this to a*sqrt 3 and solve for a.
6. have to give this one more thought. the question is a bit vague as to whether both "other" sides are sqrt2.
7.(a) cosine rule. use coordinate geometry to find the lengths of the sides and substitute into the cosine rule to find what cos ACB is.
(b) use the sine rule for area = (1/2)ab*sinC, and your newly found lengths.
8. the largest angle is the one opposite the longest side (obvious after some consideration). you can then use the cosine rule (again!), substituting in the brackets given. draw a diagram to make sure you substitute the right values in.
(b) ill leave this one to you to guess.. hint is this is a very similar format to the previous question (and q5)
(b) use the cosine rule. you now know theta, and still know the sides a and b. well, c^2 = a^2 + b^2 - 2ab*cosC.
2. Use the cosine rule, and substitute in the three lengths, a, b and c, and rearrange to find theta. alternatively, rearrange algebraically and then substitute the numbers in.
5. Use the cosine rule again. Draw a diagram, label the sides a,b,c and angles(corners) A, B and C so that angle A is opposite side a. figure out how the cosine rule will help you. it has 4 variables, and you know 3 of them. use it to find the 4th (side b, or AC). then equate this to a*sqrt 3 and solve for a.
6. have to give this one more thought. the question is a bit vague as to whether both "other" sides are sqrt2.
7.(a) cosine rule. use coordinate geometry to find the lengths of the sides and substitute into the cosine rule to find what cos ACB is.
(b) use the sine rule for area = (1/2)ab*sinC, and your newly found lengths.
8. the largest angle is the one opposite the longest side (obvious after some consideration). you can then use the cosine rule (again!), substituting in the brackets given. draw a diagram to make sure you substitute the right values in.
(b) ill leave this one to you to guess.. hint is this is a very similar format to the previous question (and q5)
*girlie*
6) In a triangle, the largest side has length 2 cm and one of the other sides has length sqrt 2 cm. Given that the area of the triangle is 1 cm^2, show that the triangle is right-angled and isosceles.
6) In a triangle, the largest side has length 2 cm and one of the other sides has length sqrt 2 cm. Given that the area of the triangle is 1 cm^2, show that the triangle is right-angled and isosceles.
(1/2)*(2)*Cos(theta) = 1
=> Cos(theta) = 1/(rt2)
=> theta = 45
By the Cosine Rule:
x^2 = (2)^2 + (rt2)^2 - 2*2*rt2*Cos45
=> x^2 = 2
=> x = rt2 (we take positive value because we can't negative length)
As both sides are root 2, and both adjacent angles are 45 degrees, the triangle is a right angeled isocoloses triangle.
Euclid
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