A boy is flying a kite with a string of length 100 m . If the string is tight and the angle of elevation of the kite is 26° 32′ , find the height of the kite correct to one decimal place ( ignore the height of the boy ) . About the author Kaylee

Answer:– the height of kite is 44.7m To Find:– the height of the kite correct to one decimal place. Given:– Angle of elevation of the kite = 26° 32′ lenght of string = 100m Consider:– AB as the height of the kite A and AC as the string, → AC = 100 → Take AB = x In right triangle ABC => Sin∅ = [tex]\frac{AB}{AC}[/tex] [tex] \:\:\:Substituting\:the\:values,[/tex] ⇒sin26° 32′ = [tex]\frac{x}{100}[/tex] ⇒0.4467 = [tex]\frac{x}{100}[/tex] ⇒x = 100 × 0.4467 ⇒x = 44.67 ≈ 44.7m Hence, the height of the kite is 44.7 m. Reply

Answer:882 is my answer. hope it will help you

Answer:–ToFind:–Given:–Consider:–## In right triangle ABC

[tex] \:\:\:Substituting\:the\:values,[/tex]

## Hence, the height of the kite is 44.7 m.