In triangle abc, ad is perpendicular to bc. Sin B = 0.8, bd = 9cm and tan C = 1. Find the length of ab, ad, ac, and dc
Answers
Answered by
39
Sin B = 0.8
So cos B = √1-cos² B = √1-(0.8)² = √1-0.64 = √0.36 = 0.6
In ΔABD,
cos B = BD/AB = 0.6 = 9/AB
AB = 9/0.6 = 15
SO AB = 15 cm
Sin B = AD/AB
0.8 = AD/15
AD = 12 cm
IN Δ ADC,
tan C= AD/CD
CD = 12 cm
and SecC = √1+ tan² C = √1 + 1² = √2
So,
SecC = AC/CD
= √2 = AC/12
= AC = 12√2 cm
So cos B = √1-cos² B = √1-(0.8)² = √1-0.64 = √0.36 = 0.6
In ΔABD,
cos B = BD/AB = 0.6 = 9/AB
AB = 9/0.6 = 15
SO AB = 15 cm
Sin B = AD/AB
0.8 = AD/15
AD = 12 cm
IN Δ ADC,
tan C= AD/CD
CD = 12 cm
and SecC = √1+ tan² C = √1 + 1² = √2
So,
SecC = AC/CD
= √2 = AC/12
= AC = 12√2 cm
Answered by
3
Step-by-step explanation:
In △ABC, AD⊥BC, sinB=0.8, tanC=1, BD=9
In △ABD,
sinB=0.8
cos
2
B=1−sin
2
B=1−(0.8)
2
cosB=(0.6)
2
cosB=0.6
cosB=
AB
BD
AB=
0.6
BD
AB=
0.6
9
=15
Also, using Pythagoras theorem,
AB
2
=AD
2
+BD
2
15
2
=9
2
+AD
2
AD=144
AD=12 cm
Now, In △ACD
tanC=1=
B
P
=
CD
AD
AD=CD=12 cm
Using Pythagoras theorem,
AD
2
+CD
2
=AC
2
12
2
+12
2
=AC
2
AC=12
2
cm
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