6)Given that cos a=5/13 and that 270°<or=a<or=360°,evaluate tan a+sin a
7)If tan titre=12/5 and titre is reflex,find the value of 5sin titre+cos titre
8)Find titre given that 2cos3titre-1=0 for 0°<or=x<or=360°
9)Solve the equation; 2cos2(x+30)°=-1 for 0°<or=x<or=360°
10)Solve for titre in the equation; cos(2titre+30)=-0.7660 for -180°<or=x<or=180°
11)Solve the equation 2 tan 2xcos2x+√3=0 for -180°<or=x<or=180°
12)Given that 3tan^2(2x+2/3π°=1,Find x when 0°<or=x<or=2π°.Give your answer in π°
13)A figure has point D which divides AC in the ratio 1:1.AB=6.4cm, angle ABD=39° and angle BAD=78°.Calculate correct to 1 d.p;
a)The length of BD
b)The length of BC
14)A triangle ABC is inscribed in a circle.AB=6cm.BC=9cm and angle ABC=80°.Calculate to one decimal place;
a)The length of AC
b)The radius of the circle
c)The area of the shaded segments
6) We are given that cos a = 5/13 and 270° < a < 360°. Since cos a = adjacent/hypotenuse, we can assume a right triangle with a = adjacent and 13 = hypotenuse. Using the Pythagorean theorem, we can find the opposite side to be 12.
Therefore, sin a = opposite/hypotenuse = 12/13
tan a = sin a / cos a = (12/13) / (5/13) = 12/5
Thus, tan a + sin a = 12/5 + 12/13 = (156 + 60)/65 = 216/65
7) We are given that tan θ = 12/5 and it is a reflex angle. Let's assume a right triangle with a = opposite, 5 = adjacent, and 13 = hypotenuse. Therefore, sin θ = 12/13 and cos θ = 5/13.
So, 5sin θ + cos θ = 5 * (12/13) + 5/13 = (60 + 5)/13 = 65/13 = 5
8) We are given that 2cos(3θ) - 1 = 0. Rearranging the equation, we get 2cos(3θ) = 1. Dividing both sides by 2, we have cos(3θ) = 1/2.
Since cos(60°) = 1/2, we can write 3θ = 60°. Solving for θ, we get θ = 60°/3 = 20°.
9) We have 2cos(2(x + 30)°) = -1. Dividing both sides by 2, we have cos(2(x + 30)°) = -1/2.
Since cos(120°) = -1/2, we can write 2(x + 30)° = 120°. Solving for x, we get x = (120° - 30°)/2 = 45°.
10) We have cos(2θ + 30) = -0.7660. Taking the inverse cosine on both sides, we get 2θ + 30 = 140.087°.
Solving for θ, we have 2θ = 140.087° - 30° = 110.087°. Dividing by 2, we get θ = 55.0435°.
11) We are given 2tan(2x)cos(2x) + √3 = 0. Rearranging, we have cos(2x)tan(2x) = -√3/2.
Since cos(60°) = 1/2 and tan(60°) = √3, we can write 2x = 60°, 2x = 120°, or 2x = 240°.
Solving for x, we have x = 30°, x = 60°, or x = 120°.
12) We have 3tan^2(2x + 2π/3) = 1. Dividing both sides by 3, we have tan^2(2x + 2π/3) = 1/3.
Taking the square root of both sides, we have tan(2x + 2π/3) = ±√(1/3).
Since tan(30°) = √(1/3), we can write 2x + 2π/3 = 30°. Solving for x, we have x = (30° - 2π/3)/2 = (90° - 2π/3)/6.
13) a) Given that AD divides AC in the ratio 1:1, we can assume that BD = CD.
Since angle ABD = 39° and angle BAD = 78°, we have angle ADB = 180° - 39° - 78° = 63°.
Using the sine rule, (AB/sin(ADB)) = (BD/sin(ABD)). Plugging in the values, we have (6.4/sin(63°)) = (BD/sin(78°)).
Solving for BD, we have BD = (6.4 * sin(78°))/sin(63°) ≈ 7.42 cm.
b) Using the angle BCD, we have angle BCD = 180° - 39° = 141°.
Using the sine rule again with triangle BCD, we have (BC/sin(BCD)) = (CD/sin(BDC)).
Plugging in the values, we have (BC/sin(141°)) = (BD/sin(39°)).
Solving for BC, we have BC = (BD * sin(141°))/sin(39°) ≈ 5.42 cm.
14) a) Using the sine rule for triangle ABC, we have (AB/sin(ACB)) = (AC/sin(ABC)).
Plugging in the values, we have (6/sin(80°)) = (AC/sin(100°)).
Solving for AC, we have AC = (6 * sin(100°))/sin(80°) ≈ 10.36 cm.
b) The center of the circle is the point where the perpendicular bisectors of the sides of the triangle intersect. Using the formula R = (AB * BC * AC)/(4 * area of triangle ABC), we have R = (6 * 9 * 10.36)/(4 * area of triangle ABC).
Using the formula area = (1/2) * AB * BC * sin(ABC), we have R = (54.36)/(4 * (1/2) * 6 * 9 * sin(80°)).
Simplifying, we have R = (54.36)/(2 * 54 * (0.9848)) ≈ 13.16 cm.
c) To find the area of the shaded segments, we need to find the area of the sector OAB and subtract the area of triangle ABC.
The angle of the sector OAB is 80°, so the area of the sector is (80/360) * π * (r^2), where r is the radius of the circle.
Using the formula for the area of a triangle, the area of triangle ABC is (1/2) * AB * BC * sin(ABC).
Therefore, the area of the shaded segments is (80/360) * π * (13.16^2) - (1/2) * 6 * 9 * sin(80°).