# Thread: trigonometry : multiple choice problems

1. ## trigonometry : multiple choice problems

1. A tower 150m high is situated at the top of a hill. At a point 650m down the hill the angle between the surface of the hill and the line of sight to the top of the tower is 12 degrees 30 minutes. Find the inclination of the hill to the horizontal plane.

Choices
a. 5 degrees 54 minutes
b. 7 degrees 10 minutes
c. 6 degrees 12 minutes
d. 7 degrees 50 minutes

Comment: With a calculator, Number 2 below seems to have no correct choices
when i substitute the equation with any angle

2. Evaluate 2tan(alpha) / (1+tan^2{alpha})
a. 2\sin{\alpha}
b.2\cos{\alpha}
c. 2\sec{\alpha}
d. 2\csc{\alpha}

3. three circles with centers A, B and C have respective radii 50, 30 and 20 inches and are tangent to each other externally. Find the area in (in^2) of the curvilinear triangle formed by the three circles
a. 142
b. 152
c. 146
d. 148

please guide me the steps in solving problem # 1 and # 3, problem 2 is easy but the answer seems not to be there

2. Hello, TechnicianEngineer!

Does #2 have any typos?

Comment: With a calculator, Number 2 below seems to have no correct choices
when i substitute the equation with any angle. . I agree!

$\displaystyle \text{2. Evaluate: }\:\frac{2\tan\alpha}{1+\tan^2\alpha}$

. . $\displaystyle (a)\;2\sin\alpha \qquad (b)\;2\cos\alpha \qquad (c)\;2\sec\alpha \qquad (d)\;2\csc\alpha$

$\displaystyle \text{We have: }\:\frac{2\tan\alpha}{1 + \tan^2\!\alpha} \;=\;\frac{2\,\dfrac{\sin\alpha}{\cos\alpha}}{1 + \dfrac{\sin^2\alpha}{\cos^2\alpha}}$

$\displaystyle \text{Multiply by }\frac{\cos^2\!\alpha}{\cos^2\!\alpha}\!:\;\;\frac {\overbrace{2\sin\alpha \cos\alpha}^{\text{This is }\sin2\alpha}}{\underbrace{\cos^2\alpha + \sin^2\alpha}_{\text{This is 1}}} \;=\; \frac{\sin2\alpha}{1} \;=\;\sin2\alpha$

3. the answer might be letter a,

pls help on the other numbers its so difficult
i dunno what's a curvilinear triangle

4. for Q.1 if A is top of tower , B is top of hill and C is the point down the hill. the for triangle ABC

AB = 150m BC = 650m <C = 12.5

using Sine law < A = 69.70

gives inclination of hill = ( 90 - 69.70 ) - 12.5 = 7.8 = 7 degree 50 min

for Q.3 sides of triangle are let a = 70 , b = 50 and c = 80 gives s = ( a + b + c ) / 2
________________
s = 100 now using formula area of triangle = / s(s-a)(s-b)(s-c) area can be calculated

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### a flagpole 160.4 meter is situated at the top of the hill. at a point 600m down the hill the angle between the surface of the hill and a line to the top of the flagpole is 8 degree. find the distance from the point to the top of the flagpole

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