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1. Triangle ABC is given where A = 42°, a = 3, and b = 8. How many distinct triangles can be made with the given measurements? Explain your answer.
2. Use basic identities to simplify the following expression.
[tex]tanx^2-sec^2x/cos x[/tex]
3.Solve the following equation for 0 ≤ ѳ < 2π .
sec ѳ2 − 6 = − 4
4. What is the exact value of cos cos 75° ?
5. Find BC. Round your answer to the nearest tenth.


1 Triangle ABC Is Given Where A 42 A 3 And B 8 How Many Distinct Triangles Can Be Made With The Given Measurements Explain Your Answer 2 Use Basic Identities To class=

Sagot :

1) Two distinct triangles can be done.

2) The result is [tex]\sec x -\cos x -\sec^{2}x[/tex] .

3) For [tex]\tan \theta = 1[/tex], the solution is [tex]\theta = \frac{\pi}{4}+ \pi\cdot i[/tex], [tex]i\in \mathbb{Z}[/tex]. For [tex]\tan \theta = -1[/tex], the solution is [tex]\theta = \frac{3\pi}{4}+\pi\cdot i[/tex], [tex]i\in \mathbb{Z}[/tex].

4) The exact value of [tex]\cos 75^{\circ}[/tex] is [tex]\frac{\sqrt{6}-\sqrt{2}}{4}[/tex].

5) The length of [tex]BC[/tex] is approximately 33.009 units.

Trigonometry issues

1) The angle [tex]A[/tex] is always opposite to side [tex]a[/tex]. A triangle is formed by three angles and three sides. In this case, an angle and two sides are known and thus we conclude that two distinct triangles can be done.

2) We proceed to apply trigonometric expressions to simplify the given expression:

i) [tex]\tan^{2}x-\frac{\sec^{2}x}{\cos x}[/tex]     Given

ii) [tex]\frac{\sin^{2}x}{\cos^{2}x}-\frac{1}{\cos^{3} x}[/tex]     [tex]\tan x = \frac{\sin x}{\cos x}[/tex], [tex]\sec x = \frac{1}{\cos x}[/tex]

iii) [tex]\frac{\sin^{2}x\cdot \cos x-1}{\cos^{2}x}[/tex]     [tex]\frac{x}{y} +\frac{w}{z} = \frac{x\cdot z+w\cdot y}{y\cdot z}[/tex]

iv) [tex]\frac{(1-\cos^{2}x)\cdot \cos x-1}{\cos ^{2}x}[/tex]     [tex]\sin^{2}x + \cos^{2}x = 1[/tex]

v) [tex]\frac{\cos x-\cos^{3}x-1}{\cos^{2}x}[/tex]     Distributive property/[tex]x^{m+n} = x^{m}\cdot x^{n}[/tex]

vi) [tex]\frac{1}{\cos x}-\cos x -\frac{1}{\cos^{2}x}[/tex]     [tex]\frac{a\cdot c}{b\cdot c} = \frac{a}{b}[/tex]/[tex]\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}[/tex]

vii) [tex]\sec x -\cos x -\sec^{2}x[/tex]     [tex]\sec x = \frac{1}{\cos x}[/tex]/Result

3) The given equation is [tex]\sec^{2}\theta - 6 = -4[/tex]. Now we proceed to simplify the formula in terms of tangents:

[tex]\tan^{2}\theta -5 = -4[/tex]

[tex]\tan^{2}\theta = 1[/tex]

[tex]\tan \theta = \pm 1[/tex]

For [tex]\tan \theta = 1[/tex], the solution is [tex]\theta = \frac{\pi}{4}+ \pi\cdot i[/tex], [tex]i\in \mathbb{Z}[/tex]. For [tex]\tan \theta = -1[/tex], the solution is [tex]\theta = \frac{3\pi}{4}+\pi\cdot i[/tex], [tex]i\in \mathbb{Z}[/tex]. [tex]\blacksquare[/tex]

4) The exact value can be found by using the following formula for the addition of two angles:

[tex]\cos (\alpha+\beta) = \cos \alpha\cdot \cos \beta - \sin \alpha \cdot \sin \beta[/tex]  (1)

If we know that [tex]\alpha = 30^{\circ}[/tex] and [tex]\beta = 45^{\circ}[/tex], then the formula is:

[tex]\cos 75^{\circ} = \cos 30^{\circ}\cdot \cos 45^{\circ}-\sin 30^{\circ}\cdot \sin 45^{\circ}[/tex]

[tex]\cos 75^{\circ} = \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2} -\frac{1}{2}\cdot \frac{\sqrt{2}}{2}[/tex]

[tex]\cos 75^{\circ} = \frac{\sqrt{6}}{4} -\frac{\sqrt{2}}{4}[/tex]

[tex]\cos 75^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}[/tex]

The exact value of [tex]\cos 75^{\circ}[/tex] is [tex]\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]. [tex]\blacksquare[/tex]

5) The length of [tex]BC[/tex] is found by cosine law:

[tex]BC = \sqrt{AC^{2}+AB^{2}-2\cdot AC\cdot AB\cdot \cos A}[/tex] (2)

[tex]BC = \sqrt{17^{2}+28^{2}-2\cdot (17)\cdot (28)\cdot \cos 91^{\circ}}[/tex]

[tex]BC \approx 33.009[/tex]

The length of [tex]BC[/tex] is approximately 33.009 units. [tex]\blacksquare[/tex]

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