Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\rm :\longmapsto\:\triangle ABC \: \sim \: \triangle DEF[/tex]
[tex]\bf\implies \:\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} [/tex]
and
[tex]\rm\implies \:\angle A = \angle D[/tex]
[tex]\rm\implies \:\angle B = \angle E[/tex]
[tex]\rm\implies \:\angle C = \angle F[/tex]
Also, given that,
[tex]\rm\implies \:\dfrac{DE}{AB} = \dfrac{DF}{AC} = \dfrac{EF}{BC} = \dfrac{1}{3} [/tex]
[tex]\rm :\longmapsto\:BC = 16 \: units[/tex]
[tex]\rm :\longmapsto\:AC = 20 \: units[/tex]
So, using Pythagoras Theorem, we have
[tex]\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2} [/tex]
[tex]\rm :\longmapsto\: {20}^{2} = {AB}^{2} + {16}^{2} [/tex]
[tex]\rm :\longmapsto\: 400 = {AB}^{2} + 256[/tex]
[tex]\rm :\longmapsto\: {AB}^{2} = 400 - 256[/tex]
[tex]\rm :\longmapsto\: {AB}^{2} = 144[/tex]
[tex]\bf\implies \:AB = 12 \: units[/tex]
Now,
[tex]\rm :\longmapsto\:sinC = \dfrac{AB}{AC} [/tex]
[tex]\rm :\longmapsto\:sinC = \dfrac{12}{20} [/tex]
[tex]\rm :\longmapsto\:sinC = \dfrac{3}{5} [/tex]
Now, As we have
[tex]\rm :\longmapsto\:\angle C = \angle F[/tex]
[tex]\rm :\longmapsto\:sin C = sin F[/tex]
[tex]\bf\implies \:sinF = \dfrac{3}{5} [/tex]
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1. Pythagoras Theorem :-
This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.
2. Converse of Pythagoras Theorem :-
This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.
3. Area Ratio Theorem :-
This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.
4. Basic Proportionality Theorem :-
This theorem states that :- If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.
