Maths Topic 4: Trigonometry Trivia Quiz

Answer: 3sqrt(5)/2

To solve this question, we first need to work out the length of the third side, which can be done through Pythagoras.
So since sinx = 2/7, the opposite side is 2, and the hypotenuse is 7
Therefore,
7^2-2^2=45
Therefore the adjacent side length is sqrt(45), or 3sqrt(5).
Now cotx is adjacent/opposite
So, cotx= 3sqrt(5)/2

Answer: secx

To solve this question, we can use a right-angled triangle with acute angles x and 90-x. For the angle 90-x, the adjacent side is equal to b, the opposite side is equal to a, and the hypotenuse is equal to c. Since the cosecant ratio is hypotenuse/opposite, therefore cosec(90-x) is equal to c/a. The secant ratio is hypotenuse/adjacent. Looking at the angle x, secx is equal to c/a.
Therefore, secx = cosec(90-x).

Answer: 25 degrees, 55 minutes

This is a question that can be solved on a calculator
So, tanx = 0.486
x = arctan(0.486)
x = 25.9197519
Therefore, x = 25 degrees, 55 minutes.

Answer: 14 cm

From the information given, we can use the cosine ratio to solve this question.
So,
cos(39d,49m)= x/18.2
x = 18.2cos(39d,49m)
x = 13.97937069
Therefore, x = 14cm, rounded up.

Answer: 25 degrees, 16 minutes

From the information given, we can use the sine ratio to solve this question.
So,
sinx = 3.5/8.2
x = arcsin(3.5/8.2)
x = 25.26650518
Therefore, x = 25 degrees, 16 minutes.

Answer: 14.63m

Using the information given, we can draw a right-angled triangle, with hypotenuse 20m, an acute angle of 47, and the opposite side of the angle to be x.
So,
sin47 = x/20
x = 20sin47
x = 14.62707403
Therefore, x = 14.63m

Answer: 310 degrees

Bearings are measured starting from a north line in a clockwise direction. From the information given, we can see that point B is south-east of point X. To find the bearing of X from B, we would need to find the reflex angle at B.
Using the north line at B, and co-interior angles,
180 - 130 = 50
Therefore
360 - 50 = 310

Therefore, the bearing of X from B is 310 degrees.

Answer: 1.25

To solve this question, we can use exact values for sin60 and cos45
So, sin60=sqrt(3)/2, cos45=1/sqrt(2)
Therefore,
(sqrt(3)/2)^2+(1/sqrt(2))^2
= 0.75+0.5
= 1.25

Answer: -3/4

From the information given, we can work out that the remaining side is equal to 4. We can also see that since sinx is negative and cosx is positive, that the answer lies in the 4th quadrant. This makes tanx negative.
Therefore, tanx = -3/4.

Answer: 45, 135, 225, 315

2sin^2(x)= 1
sin^2(x)= 1/2
sin(x) = +-(1/sqrt(2))
Basic angle = 45
Angle lies in all quadrants
Therefore,
x = 45, 180-45, 180+45, 360-45
x = 45, 135, 225, 315

Answer: 1/(1+cosx)

We can use trigonometric identities, to find the correct expression.
So, using sin^2(x)+cos^2(x)= 1
(1-cosx)/sin^2(x)
=(1-cosx)/(1-cos^2(x))
=(1-cosx)/((1-cosx)(1+cosx))
= 1/(1+cosx)
Therefore, (1-cosx)/sin^2(x)= 1/(1+cosx).

Answer: 10cm

To solve this question, we need to use the sine rule,
which is, a/sinA = b/sinB = c/sinC.
So,
XZ/sin103 = 8/sin53
XZ = 8sin103/sin53
XZ = 9.760348019
Therefore, XZ is equal to 10cm.

Answer: 9.99 cm

To solve this question, we need to use the cosine rule,
which is, a^2=b^2+c^2-2bccosA.
So,
AC^2 = 5.6^2+6.4^2-2(5.6)(6.4)cos(112d,32m)
AC^2 = 99.78927117
AC = 9.989458002
Therefore, AC equals 9.99cm.

Answer: 11.96 units^2

We can solve this question by using the formula for area, A=(1/2)absinC.
So,
A =(1/2)(4.3)(5.8)sin(106d,22m)
A = 11.96469156
Therefore, the area of the triangle is 11.96 units^2.

Answer: 160.3 m

If we let the top of the hill be point H, then from the information given, we can see that ABH is a right-angled triangle, with angle HBA equal to 27, and AB equal to 500.
So,
Tan27 = AH/500
AH = 500tan27
Now, if we let the height of the hill equal h
Then, sin39 = h/AH
h = AHsin39
h = 500tan(27)sin(39)
h = 160.3273776
Therefore, the height of the hill is 160.3m.

Answer: (sqrt(6)-sqrt(2))/4

To solve this question, we need to use the formula cos(A+B)=cosAcosB-sinAsinB.
So,
cos75
=cos(45+30)
=cos45cos30-sin45sin30
Then using exact values,
=(1/sqrt(2))(sqrt(3)/2)-(1/sqrt(2))(1/2)
= sqrt(3)/2sqrt(2)-1/2sqrt(2)
=(sqrt(3)-1)/2sqrt(2) x sqrt(2)/sqrt(2)
=(sqrt(6)-sqrt(2))/4

Answer: 1/sqrt(3)

To solve this question, we need to use the rule,
Tan(2A)=2TanA/(1-Tan^2(A))
So,
2tan15/(1-tan^2(15))
= tan(2(15))
= tan30
= 1/sqrt(3)
Therefore, the exact value of 2tan15/(1-tan^2(15)) is 1/sqrt(3).

Answer: (1-t^2)/(t^2+1)

To solve this question, we first need to make a triangle with x/2 as an acute angle, and t and 1 as the opposite and adjacent sides respectively. Using this, we can work out the hypotenuse to be sqrt(t^2+1).
Therefore,
sin(x/2)= t/sqrt(t^2+1)
cos(x/2)= 1/sqrt(t^2+1)
Now, we can use the rule cos2x = cos^2(x)-sin^2(x)
so, cosx = cos^2(x/2)-sin^2(x/2)
cosx =(1/sqrt(t^2+1))^2-(t/sqrt(t^2+1))^2
cosx = 1/(t^2+1)-t^2/(t^2+1)
Therefore, cosx = (1-t^2)/(t^2+1).

Answer: 60, 150, 240, 330

To solve this question, we first need to expand the left hand side of the equation.
So,
Tan(x+60)= -tanx
(tanx+tan60)/(1-tan(x)tan(60))= -tanx
(tanx+sqrt(3))/(1-sqrt(3)tan(x))= -tanx
tanx+sqrt(3) = -tanx+sqrt(3)tan^2(x)
sqrt(3)tan^2(x)-2tanx-sqrt(3)= 0
This can then be solved like a quadratic
(sqrt(3)tanx+1)(tanx-sqrt(3))= 0
tanx = -1/sqrt(3), tanx=sqrt(3)
For the first solution,
basic angle = 30
angle lies in 2nd, 4th quadrants
Therefore, x = 180-30, 360-30
x = 150, 330
For the second solution,
basic angle = 60
angle lies in 1st, 3rd quadrants
Therefore, x = 60, 180+60
x = 60, 240

Answer: x = 180n+45

The general solution of equation involving tan is x = 180n+a, where a is the basic angle from the given equation. In this case, a = 45, therefore, x = 180n+45.