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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,608

T # 1

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

that's easy.

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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
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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.

Character is who you are when no one is looking.

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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: 16,608

Good, mathsyperson.

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
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T # 2

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

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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1/9

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,608

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I was just guessing.

:D:D

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,608

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.

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

ANyway, give me another for guessing.

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,608

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.

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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: 16,608

Excellent! Good work, mathsyperson!

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,608

T # 4

Show that

Cos^4A - Cos²A = Sin^4A-Sin²A

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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: 16,608

You're correct, mathsyperson!

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,608

T # 5

Find the value of Sin 15°

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

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

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**ganesh****Moderator**- Registered: 2005-06-28
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T # 7

Solve:-

Cos²θ/(Cot²θ-Cos²θ) = 3.

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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.

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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
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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.

Character is who you are when no one is looking.

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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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