Question

In triangle abcabca, b, c, the measure of \angle b∠bangle, b is 90^\circ90 ​∘ ​​ 90, degree, bc=16bc=16b, c, equals, 16, and ac=20ac=20a, c, equals, 20. triangle defdef d, e, f is similar to triangle abcabca, b, c, where vertices ddd, eee, and fff correspond to vertices aaa, bbb, and ccc, respectively, and each side of triangle def d, e, f is \dfrac{1}{3} ​3 ​ ​1 ​​ start fraction, 1, divided by, 3, end fraction the length of the corresponding side of triangle abcabca, b,
c. what is the value of \text{sin }f sin fs, i, n, space, f

Answer 1

we know that

ABC is a right triangle

m∠B=90°

AC=20

BC=16

AB=?

see the attached figure to better understand the problem

Step 1

Applying the Pythagorean Theorem

Find the value of AB

$AC^{2} =AB^{2} +BC^{2}\\ AB^{2}=AC^{2}- BC^{2} \\AB^{2}=20^{2}- 16^{2} \\AB^{2}=20^{2}- 16^{2} \\AB=12\ units$

Step 2

we know that

triangle ABC and triangle DEF are similar and the scale factor  is equal to (1/3)

Find the measures of the triangle DEF

$DE=AB*(1/3)=(12/3)\ units \\EF=BC*(1/3)=(16/3)\ units\\DF=AC*(1/3)=(20/3)\ units$

Step 3

Find the value of sin F

we know that

In a right triangle, the value of the sine is equal to

$sin\ F= \frac{opposite\ side\ angle\ F}{hypotenuse}$

in this problem

$opposite\ side\ angle\ F=DE=(12/3) units\\ hypotenuse=DF=(20/3) units$

substitute

$sin\ F= \frac{(12/3)}{(20/3)}$

$sin\ F= \frac{(12)}{(20)}$

$sin\ F= \frac{3}{5}$

therefore

the answer is

the value of sin F is equal to $\frac{3}{5}$