Question

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

Answer 1

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°