Ex 5.5
1 answer:
;
2 × 5 × sin( π / 12 ) = 10 × sin ( π / 12 ) ;
π/12 = ( 180° ) / 12 = 15°.
<span>
But, sin ( π/12 ) </span>
<span>= sin 15° </span>
<span>= sin ( 45° - 30°) </span>
<span>= sin 45°· cos 30° - cos 45°· sin 30° </span>
<span>= (1/√2)·(√3 /2 ) - (1/√2)·(1/2) </span>
<span>= ( √3 - 1 ) / (2√2) = </span>
;
Finally, ( 2.5 ) × ( 2.4 - 1.4 ) = 2.5 cm ;
2 × 5 × sin( π / 12 ) = 10 × sin ( π / 12 ) ;
π/12 = ( 180° ) / 12 = 15°.
<span>
But, sin ( π/12 ) </span>
<span>= sin 15° </span>
<span>= sin ( 45° - 30°) </span>
<span>= sin 45°· cos 30° - cos 45°· sin 30° </span>
<span>= (1/√2)·(√3 /2 ) - (1/√2)·(1/2) </span>
<span>= ( √3 - 1 ) / (2√2) = </span>
Finally, ( 2.5 ) × ( 2.4 - 1.4 ) = 2.5 cm ;
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Answer:
Step-by-step explanation:
Given: △ABC, BC>AC, D∈ AC , CD=CB
To prove: m∠ABD is acute
Proof: In ΔABC, the angle opposite to side BC is ∠BAC and the angle opposite to side AC is ∠ABC.
Now, it is given that BC>AC, then ∠BAC>∠ABC.. (1)
In ΔBDC, using the exterior angle property,
∠ADB=∠DBC+∠BCD
∠ADB=∠DBC+∠BCA
⇒∠ADB>∠BAC (2)
From equation (1) and (2), we get
∠ADB>∠BAC
⇒∠ADB>∠ABC
⇒DB>AB
Hence, m∠ABD is acute
