Question

From the following figure , find the values of a) sin B b) tan C c) sec^2B-tan^2B d) sin ^2C +cos^2C​

Answer 1

Answer:

a) 12/13 b) 3/4 c) 1 d) 1

Step-by-step explanation:

In ΔABD

AD²=AB²-BD²

=13²-5²=169-25

=144

So AD=12

In the ΔACD

AC²=DC²AD²

=16²+12²

=256+144=400

AC=20

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Now a) sinB=AD/AB=12/13

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b)  tanC=AD/DC=12/16=3/4

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c) sec²B-tan²B

=(13/5)²-(12/5)²

=169/25-144/25

=(169-144)/25

=25/25=1

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d) sin²C +cos²C

=((12/20)²+(16/20)²

=144/400+256/400=400/400=1

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