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Dmitry_Shevchenko [17]
4 years ago
8

If triangle ABC, m B = 90°, cos(9 = 17, and AB = 16 units.

Mathematics
1 answer:
Fed [463]4 years ago
7 0

Answer:

m∠A = 62°

m∠C = 28°

AC = 17 units

Step-by-step explanation:

In the given triangle ABC,

m∠B = 90°, Cos(C) = \frac{15}{17} and AB = 16 units

Since, Cos(C) = \frac{\text{Corresponding side}}{\text{Hypotenuse}}

Cos(C) = \frac{\text{Corresponding side}}{\text{Hypotenuse}}=\frac{15}{17}

\text{C}=\text{Cos}^{-1}(\frac{15}{17})

m∠C = 28.07°

m∠C ≈ 28°

Therefore, side BC = 15 units and AC = 17 units

Now we apply Sine rule in the given triangle.

Sin(A) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

         = \frac{\text{BC}}{\text{AC}}

         = \frac{15}{17}

A = \text{Sin}^{-1}(\frac{15}{17} )

A = 61.93°

m∠A = 62°

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