Question

In a figure AD is perpendicular to BC ,BD =15cm,sinB=4/5 and tanC=1.find. 1) length of AD, AB, DC, AC.

Answer 1

Step-by-step explanation:

It is given that

$sin \: B= \frac{4}{5}$

As we know that sin B = AD/AB=4/5

AD/AB= 4k/5k..... eq1

tan C= AD/DC

tan C=1 (given)

AD/DC= 1

AD=DC....eq2

in right triangle ∆ABD

${AB}^{2} = {AD}^{2} + {BD}^{2}$

${(5k)}^{2} = {(4k)}^{2} + ( {15)}^{2} \\ \\ 25 {k}^{2} = 16 {k}^{2} + 225 \\ \\ 25 {k}^{2} - 16 {k}^{2} = 225 \\ \\ 9 {k}^{2} = 225 \\ \\ {k}^{2} = \frac{225}{9} \\ \\ {k}^{2} = 25 \\ \\ k = 5$

So,

AD=4k=4×5 =20 cm

AB= 5k=5×5=25 cm

DC = 20cm( from eq2)

To find AC,apply Pythagoras Theorem in ∆ACD

${AC}^{2} = {AB}^{2} + {DC}^{2} \\ \\ = 625 + 400 \\ \\ = 1025 \\ \\ ac = \sqrt{1025} \\ \\ ac = 5 \sqrt{41} \: cm$

Hope it helps you.

Answer 2

Answer:

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