In triangleABC, angle B= 90 degrees, cos(C)=15/17 and AB = 16 what is the measure of Angle A, Angle C and angle AC

m∠C=28°, m∠A=62°, AC=34.1 units

Step-by-step explanation:

Given In ΔABC, m∠B = 90°, , and AB = 16 units. we have to find m∠A, m∠C, and AC.

As,  cos(C)={15}/{17}

⇒ angle C=cos^{-1}(\frac{15}{17})=28.07^{\circ}\sim28^{\circ}

By angle sum property of triangle,

m∠A+m∠B+m∠C=180°

⇒ m∠A+90°+28°=180°

⇒ m∠A=62°

Now, we have to find the length of AC

sin 28^{\circ}=\frac{AB}{AC}

⇒ AC=\frac{16}{sin 28^{\circ}}=34.1units

The length of AC is 34.1 units

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.