Chetan and bala are standing at point A. Chetan starts running in the direction of Southwest making an angle of ‘alpha’ with the South direction and stood at point B. Bala starts running in the direction of South-east making an angle of ‘beta’ with the South direction and stood at point C. The perpendicular drawn from point A meets line segment joining the points B and C at the point D.
Answer the following
1) what is the value of cot B + cot C?
2) What is the possible value of cos C?
3) if sin alpha = 7/8, then the value of cos ² alpha is
4) if alpha = 30° and beta = 30°, then the value of sin ²alpha + cos ²beta =
5) if cos beta = 12/13, then the value of cos C is
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Answer:
cot B = BD/AD ; cot C = CD/AD
cotB + cotC
BD/AD + CD/AD = BC/AD
The answers are:
1. cot B [tex]= \frac{BD}{AD}[/tex] and cot C [tex]= \frac{CD}{AD}[/tex]
2. cos C [tex]= \frac{CD}{CA}[/tex]
3. [tex]cos^2\alpha = \frac{15}{64}[/tex]
4. [tex]sin^2\alpha + cos^2\beta = 1[/tex]
5. [tex]cos C = \frac{5}{13}[/tex]
Step-by-step explanation:
See the attached figure for a better understanding.
1. Cot or cotangent is the ratio of base length and perpendicular of the corresponding angle.
Hence, cot B [tex]=\frac{Base}{perpendicular} = \frac{BD}{AD}[/tex] and cot C [tex]=\frac{Base}{perpendicular} = \frac{CD}{AD}[/tex]
2. Cos or cosine is the ratio of base to the hypotenuse of a particular angle.
Hence, cos C [tex]=\frac{Base}{hypotenuse} = \frac{CD}{CA}[/tex]
3. Given [tex]sin\alpha = \frac{7}{8}[/tex] and to find the value of [tex]cos^2\alpha[/tex]
To solve this, we would use the identity [tex]cos^2\alpha = 1-sin^2\alpha[/tex]
Substituting the value of sin alpha in the identity, we obtain,
[tex]cos^2\alpha = 1-(\frac{7}{8})^2[/tex]
[tex]cos^2\alpha = 1-(\frac{49}{64})[/tex]
[tex]cos^2\alpha = (\frac{64-49}{64})[/tex]
[tex]cos^2\alpha = \frac{15}{64}[/tex]
4. Given: [tex]\alpha = 30[/tex]° and [tex]\beta = 30[/tex]°
To find: [tex]sin^2\alpha + cos^2\beta[/tex]
Substituting the values of the angles in the expression, we obtain,
[tex]sin^2\alpha + cos^2\beta[/tex]
[tex]= sin^230 + cos^230[/tex]
[tex]=(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2[/tex]
[tex]=\frac{1}{4} + \frac{3}{4}[/tex]
[tex]=\frac{1+3}{4}[/tex]
[tex]=\frac{4}{4}[/tex]
[tex]=1[/tex]
5. Given: [tex]cos\beta =\frac{12}{13}[/tex]
To find cos C
[tex]cos C = \frac{ base}{hypotenuse} = \frac{CD}{CA}[/tex] and also [tex]cos\beta =\frac{12}{13} = \frac{base}{hypotenuse} = \frac{AD}{CA}[/tex]
So, AD = 12 cm, and CA = 13 cm, but we need to find the value of CD
Consider triangle ACD,
[tex]CA^2 = CD^2 + AD^2[/tex] ( Pythagoras Theorem)
[tex]13^2 = CD^2 + 12^2[/tex]
[tex]CD^2 = 13^2 - 12^2[/tex]
[tex]CD^2 = 169 - 144 = 25[/tex]
[tex]CD = 5[/tex]
Hence, [tex]cos C = \frac{ base}{hypotenuse} = \frac{CD}{CA}[/tex] [tex]= \frac{5}{13}[/tex]