Question

In the given Fig. 8.42, ZACB = 90°, ZBDC = 90° CD = 4 cm, BD= 3 cm, AC=12 cm. cos 4-sin A is equal to 5 12 (b 13 T (C) (0 12​

Answer 1

Answer:

BDC=90∘

∠ABC=90∘

Let ∠BCD=x∘

Using triangle sum property in △BDC, ∠DBC=90−x∘

Also ∠ABD=x∘

Using triangle sum property in △ADB, ∠BAD=90−x∘

Now considering △BDC and △ADB

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]So by AA

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]So by AA△BDC∼△ADB

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]So by AA△BDC∼△ADBHence CDBD=BDAD

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]So by AA△BDC∼△ADBHence CDBD=BDADCD8=84

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]So by AA△BDC∼△ADBHence CDBD=BDADCD8=84CD=16

BDC=∠BDA              [∵∠BDC=∠BDA=90∘]∠DBC=∠DAB              [∵∠DBC=∠DAB=(90−x)∘]So by AA△BDC∼△ADBHence CDBD=BDADCD8=84CD=16Hence CD=16cm

Step-by-step explanation:

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