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kap26 [50]
3 years ago
12

Find sin A and sec B given a=2 b=11

Mathematics
2 answers:
zalisa [80]3 years ago
8 0

Answer:

\sin A=\dfrac{2}{5\sqrt5}

\sec B=\dfrac{5\sqrt5}{11}

Step-by-step explanation:

Given: a=2,b=11

Using pythagoreous theorem:

c^2=a^2+b^2

c^2=11^2+2^2

c=5\sqrt{5}

Trigonometric Identities:

\sin \theta=\dfrac{\text{Opposite}}{\text{Hypotenuse}}

\sin A=\dfrac{a}{c}

\sin A=\dfrac{2}{5\sqrt5}

\sec B=\dfrac{c}{b}

\sec B=\dfrac{5\sqrt5}{11}

Hence, The value of trigonometric identity

\sin A=\dfrac{2}{5\sqrt5}

\sec B=\dfrac{5\sqrt5}{11}

Sonbull [250]3 years ago
7 0
If there is a triangle with Sid a and b then sina b/✓a2+b2and seven=✓a2+b2/a
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