Mathematics: Trevor, robert, and maurice made some paper airplanes and want to see how far they will fly. the distance from trevor to robert is 13.5 feet. the distance from robert to maurice is 14.2 feet. the distance from maurice to trevor is 6.6 feet. find the measures of the three angles in the triangle.a. m∠t=82.2, m∠r=27.4, m∠m=70.4b. m∠t=27.4, m∠r=70.4, m∠m=82.2c. m∠t=70.4, m∠r=82.2, m∠m=27.4d. m∠t=60, m∠r=60, m∠m=60

2x^2-3x+2=02[x^2-3/2x]+2=02[(x-3/4x)^2-9/16]+2=02(x-3/4x)^2-9/8+2=02(x-3/4x)^2+7/8=02(x-3/4x)^2=-7/8(x-3/4x)^2=-7/16x-3/4x=√7/4ix=√7/4i+3/4 x=(3+√7i)/4...

Trevor, robert, and maurice made some paper airplanes

Ответ:

alejdnn889283

06.01.2022

Mathematics

a. m∠T=82.2, m∠R=27.4, m∠M=70.4

Step-by-step explanation:

Let T, R and M represent the position of Trevor, Robert, and Maurice respectively,

Given,

TR = 13.5 feet,

RM = 14.2 feet,

TM = 6.6 feet,

Since, TRM forms a triangle, ( because sum of any two sides is greater than third side )

By the law of cosine,

$TR^2=RM^2+TM^2-2\times RM\times TM\times cos M$

$\implies cos M=\frac{RM^2+TM^2-TR^2}{2\times RM\times TM}$

By substituting the values,

$cos M=\frac{14.2^2+6.6^2-13.5^2}{2\times 14.2\times 6.6}$

$cos M=\frac{62.95}{187.44}$

$\implies m\angle M=cos^{-1}(\frac{62.95}{187.44})=70.3763251924\approx 70.4^{\circ}$

Similarly,

$cos R=\frac{TR^2+RM^2-TM^2}{2\times TR\times RM}$

$=\frac{13.5^2+14.2^2-6.6^2}{2\times 13.5\times 14.2}$

$=\frac{340.33}{383.4}$

$\implies m\angle R=cos^{-1}(\frac{340.33}{383.4})=27.4189643117\approx 27.4^{\circ}$

$cos T=\frac{TR^2+TM^2-RM^2}{2\times TR\times TM}$

$=\frac{13.5^2+6.6^2-14.2^2}{2\times 13.5\times 6.6}$

$=\frac{24.17}{178.2}$

$\implies m\angle T=cos^{-1}(\frac{24.17}{178.2})=82.2047104959\approx 82.2^{\circ}$

Hence, option 'a' is correct.

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