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Nina [5.8K]
3 years ago
13

Cos(a) 63/65 to find sin(a) and tan(a)

Mathematics
2 answers:
wlad13 [49]3 years ago
4 0

Answer:

Sin(a) = 16/65

Tan(a) = 16/63

Step-by-step explanation:

Cos(a) = adj/hyp

opp² = 65² - 63²

Opp = 16

Sin(a) = 16/65

Tan(a) = 16/63

Assuming a is acute.

If angle is in the fourth quadrant, sin and tan will be negative

Semenov [28]3 years ago
3 0

Answer:

\huge\boxed{\sin\alpha=-\dfrac{16}{65},\ \tan\alpha=-\dfrac{16}{63}}\\\\or\\\\\huge\boxed{\sin\alpha=\dfrac{16}{65},\ \tan\alpha=\dfrac{16}{63}}

Step-by-step explanation:

\cos\alpha=\dfrac{63}{65}\\\\\text{use}\ \sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\alpha+\left(\dfrac{63}{65}\right)^2=1\\\\\sin^2\alpha+\dfrac{3969}{4225}=1\qquad\text{subtract}\ \dfrac{3969}{4225}\ \text{from both sides}\\\\\sin^2\alpha=\dfrac{4225}{4225}-\dfrac{3969}{4225}\\\\\sin^2\alpha=\dfrac{256}{4225}\to\sin\alpha=\pm\sqrt{\dfrac{256}{4225}}\\\\\sin\alpha=\pm\dfrac{\sqrt{256}}{\sqrt{4225}}\\\\\sin\alpha=\pm\dfrac{16}{65}

\text{use}\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\\\\\text{substitute:}\\\\\tan\alpha=\dfrac{\pm\frac{16}{65}}{\frac{63}{65}}=\pm\dfrac{16}{65}\cdot\dfrac{65}{63}=\pm\dfrac{16}{63}

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Answer:

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Similar triangles: small & medium

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Large                  ? → 40 m          ?                       32+18 = 50 m
Medium            32 m                     24 m                 ? → 40 m
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medium and large: long leg / hypotenuse

32/40 = x/50
32*50 = 40 * x
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1600/40 = x
40 = x
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