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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

T # 1

For θ=30°, prove that tan² θ+1 = sec²θ.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

that's easy.

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

It's a well-known identity that tan²θ+1 ≡ sec²θ, so the equation is true, regardless of the value of θ.

I like how you've branched the puzzles into different categories though. ^_^

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

Thanks, mathsyperson.

Although tan² θ+1 = sec²θ is a well-known identity, this question was given to check whether the candidate remembers and uses in calculation the value of sinθ, cosθ, tanθ, secθ etc. for θ=30 degrees.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Ah. OK then.

All I know is that sin 30θ = 1/2 and that cosθ is √3/2, but I can work everything out from that.

tanθ = sinθ ÷ cosθ = (1/2)/(√3/2) = 1/√3

Therefore, tan²θ = 1/3.

secθ = 1 ÷ cosθ = 1/(√3/2) = 2/√3

Therefore, sec²θ = 4/3.

Subsituting these into the given equation gives 1/3 + 1 = 4/3, which is obviously true.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

Good, mathsyperson.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

T # 2

What is the value of (1-CosA) / (1+CosA) given that tan A = 3/4?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

1/9

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I was just guessing.

:D:D

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

krassi_holmz, what a guess!

Anyhow, this is how it's solved.

tanA = Opposite side/Adjacent side=3/4,

Therefore, Hypotenuse=[(3²+4²)]⊃½=5,

therefore, SinA=3/5, CosA=4/5,

(1-CosA)/((1+CosA) = (1/5)/(9/5) = 1/9.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

ANyway, give me another for guessing.

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

That was a great guess, krassi_holmz!

T # 3

From the top and bottom of a tower, the angle of elevation of the top of a cliff of height 400m are observed to be 30 degrees and 60 degrees.Determine the height of the tower.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The angle at the bottom of the tower is 60°.

Trigonometry can be used to show that the tower is 400/tan 60° = 400/√3m away from the cliff.

Using this information, we can then use trigonometry again to show that for the angle to be 30° at the top of the tower, the cliff must be (400/√3)/√3 = 400/3m higher than the tower.

Therefore, the tower is 400 - 400/3 = 266.67m high.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

Excellent! Good work, mathsyperson!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

T # 4

Show that

Cos^4A - Cos²A = Sin^4A-Sin²A

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Factorising both sides gives:

cos²A(1-cos²A) = sin²A(1-sin²A)

However, using the identity that sin²θ + cos²θ = 1, we can replace (1-cos²A) with sin²A and (1-sin²A) with cos²A.

Therefore, both sides equal cos²Asin²A.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

You're correct, mathsyperson!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

T # 5

Find the value of Sin 15°

(Given Sin45°=1/√2, Cos45°=1/√2, Sin30°=1/2, Cos30°=√3/2)

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

T # 6

A 25m high vertical tower and a vertical pole stand on an inclined ground. The foot of the pole and the midpoint of the pole are in the same horizontal level. The angles of depression of the top and bottom of the pole as seen from the top of the tower are 15° and 45° respectively. Find the length of the pole.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

T # 7

Solve:-

Cos²θ/(Cot²θ-Cos²θ) = 3.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
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T # 8

If x*sin60°*tan30°=(sin45°*cot45°)/(cos30°*cosec60°), find x.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

i would rather use a calculator for this prob.

**X'(y-Xβ)=0**

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

George, y,

It is helpful in trignometry to know the values of sin, cos and tan of 0°, 30°, 45°, 60°, and 90°. For other angles, the tables can be referred or a calculator can be used.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**fusilli_jerry89****Member**- Registered: 2006-06-23
- Posts: 86

ganesh wrote:

T # 7

Solve:-

Cos²θ/(Cot²θ-Cos²θ) = 3.

Not too sure how to do this but i manipulated the question until i got down to 4cos sin = 3 and then sin ≠ 3 and cos =3/4 I just dont know how to find this without a calculator.

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