Trigonometric Problem...need a solution

Johnny Blaze

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Aug 8, 2019
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In triangle ABC right angle at A. If Sin B Sin C = 0.2 and Sin (B - C) = 5a/2. Find the value of 5a - 1. Please help me. Thanks.

Here's my working

Since angle A = 90
Angle (B + C) = 90

Sin (B + C) = Sin 90 = 1
Cos (B + C) = Cos 90 = 0

Sin (B + C) = Sin B Cos C + Cos B Sin C
Cos (B + C) = Cos B Cos C - Sin B Sin C
0 = Cos B Cos C - 0.2
Cos B Cos C = 0.2
2 Cos B Cos C = 0.4
Cos (B + C) + Cos (B - C) = 0.4
0 + Cos (B - C) = 0.4
Cos (B - C) = 0.4 = 2/5

Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2
Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5

Now I'm really stuck here. Please help!

 

Jomo

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I am not sure why you are looking at trig functions for the angle B+C

Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2
Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5
You know Sin B Sin C = 0.2 So Cos B Cos C=2/5 -0.2 = 0.2
Please answer these two questions (I would draw the triangle 1st!)
1) Cos(?) = Sin C
2) Cos(?) = SinB
I would also use Sin(-x) = - Sin(x)

Show us what you can do from here.
 

Johnny Blaze

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I am not sure why you are looking at trig functions for the angle B+C

Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2
Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5
You know Sin B Sin C = 0.2 So Cos B Cos C=2/5 -0.2 = 0.2
Please answer these two questions (I would draw the triangle 1st!)
1) Cos(?) = Sin C
2) Cos(?) = SinB
I would also use Sin(-x) = - Sin(x)

Show us what you can do from here.
1) Cos (90 - B) = Sin C
2) Cos (90 - C) = Sin B
 

Jomo

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Subhotosh Khan

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Jomo - Are you sure?

If those answers are true then

Cos(90-B) = Sin(B) .... universally true \(\displaystyle \to\) Sin(B) = Sin(C) \(\displaystyle \to\) B = C ............or......... B = \(\displaystyle \pi\) - C

Similar conclusion from "Cos(90-C) = Sin(B)"

Yes, but what is 90-B? Is it A, B, C or something else? What about 90-C?
 

Jomo

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Jomo - Are you sure?

If those answers are true then

Cos(90-B) = Sin(B) .... universally true \(\displaystyle \to\) Sin(B) = Sin(C) \(\displaystyle \to\) B = C ............or......... B = \(\displaystyle \pi\) - C

Similar conclusion from "Cos(90-C) = Sin(B)"
I better get a teaching job soon as I am losing my ability to do basic math.
In a right triangle ABC with A being the right angle we have: Sin(B) = Cos(C) and Sin(C)=Cos(B)
 

Johnny Blaze

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Those are INCORRECT!

\(\displaystyle Cos(\frac{\pi}{2} - \theta) = Sin(\theta)\)
Yes, I was wrong. It should be

Since A is right-angle,
Cos (90 - B) = Sin B
Cos (90 - C) = Sin C
 

Subhotosh Khan

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Yes, I was wrong. It should be

Since A is right-angle,
Cos (90 - B) = Sin B
Cos (90 - C) = Sin C
Now continue..... (review the responses above with paper&pencil)
 

Johnny Blaze

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Aug 8, 2019
Messages
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In triangle ABC right angle at A. If Sin B Sin C = 0.2 and Sin (B - C) = 5a/2. Find the value of 5a - 1. Please help me. Thanks.

Here's my working

Since angle A = 90
Angle (B + C) = 90

Sin (B + C) = Sin 90 = 1
Cos (B + C) = Cos 90 = 0

Sin (B + C) = Sin B Cos C + Cos B Sin C
Cos (B + C) = Cos B Cos C - Sin B Sin C
0 = Cos B Cos C - 0.2
Cos B Cos C = 0.2
2 Cos B Cos C = 0.4
Cos (B + C) + Cos (B - C) = 0.4
0 + Cos (B - C) = 0.4
Cos (B - C) = 0.4 = 2/5

Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2
Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5

Now I'm really stuck here. Please help!

Since angle A = 90
Angle (B + C) = 90

Sin (B + C) = Sin 90 = 1
Cos (B + C) = Cos 90 = 0

Sin (B + C) = Sin B Cos C + Cos B Sin C
Cos (B + C) = Cos B Cos C - Sin B Sin C
0 = Cos B Cos C - 0.2
Cos B Cos C = 0.2
2 Cos B Cos C = 0.4
Cos (B + C) + Cos (B - C) = 0.4
0 + Cos (B - C) = 0.4
Cos (B - C) = 0.4 = 2/5

Sin (B - C) = 5a/2

Since,
Cos ^2 (B - C) + Sin ^2 (B - C) = 1
(2/5)^2 + (5a/2)^2 = 1
4/25 + 25a^2/4 = 1
16/100 + 625 a^2/100 = 100/100
625 a^2 = 84
a^2 = 84/625
a = 2✓21/25

5a - 1 = 5 (2✓21/25) - 1
= 2✓21/5 - 1

Is it correct?
 

Jomo

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Looks good to me
 

hoosie

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