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solve the triangle. a = 13.0 ft, b = 14.0 ft, c = 26.0 ft what is the m…

Question

solve the triangle.
a = 13.0 ft, b = 14.0 ft, c = 26.0 ft

what is the measure of angle a?
a = \\(\square\\)°
(round to two decimal places as needed.)
what is the measure of angle b?
b = \\(\square\\)°
(round to two decimal places as needed.)
what is the measure of angle c?
c = \\(\square\\)°
(round to two decimal places as needed.)

Explanation:

Step1: Use Law of Cosines for angle A

The Law of Cosines formula for angle \( A \) is \( \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} \). Substitute \( a = 13.0 \), \( b = 14.0 \), \( c = 26.0 \):

$$ \cos A=\frac{14.0^{2}+26.0^{2}-13.0^{2}}{2\times14.0\times26.0}=\frac{196 + 676-169}{728}=\frac{703}{728}\approx0.96566 $$

Then \( A=\arccos(0.96566)\approx15.07^{\circ} \)

Step2: Use Law of Cosines for angle B

The formula for angle \( B \) is \( \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac} \). Substitute the values:

$$ \cos B=\frac{13.0^{2}+26.0^{2}-14.0^{2}}{2\times13.0\times26.0}=\frac{169 + 676 - 196}{676}=\frac{649}{676}\approx0.96006 $$

Then \( B=\arccos(0.96006)\approx15.83^{\circ} \)

Step3: Use angle sum property for angle C

Since the sum of angles in a triangle is \( 180^{\circ} \), \( C = 180^{\circ}-A - B \). Substitute \( A\approx15.07^{\circ} \), \( B\approx15.83^{\circ} \):

$$ C=180-(15.07 + 15.83)=149.10^{\circ} $$

Answer:

\( A\approx15.07^{\circ} \)
\( B\approx15.83^{\circ} \)
\( C\approx149.10^{\circ} \)