Select the correct answer.
1 answer:
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Given :
- ZC = 90°
- CD is the altitude to AB.
A = 65°.
To find :
- the angles in △CBD and △CAD if m∠A = 65°
Solution :
In Right angle △ABC,
we have,
=> ACB = 90°
=> CAB = 65°.
So,
=> ACB +
CAB+
ZCBA = 180° (By angle sum Property.)
=> 90° + 65° + CBA = 180°
=> 155° +CBA = 180°
=> CBA = 180° - 155°
=> CBA = 25°.
In △CDB,
=> CD is the altitude to AB.
So,
=> CDB = 90°
=> CBD =
CBA = 25°.
So,
=> CBD +
DCB = 180° (Angle sum Property.)
=> 90° +25° + DCB = 180°
=> 115° + DCB = 180°
=> DCB = 180° - 115°
=> DCB = 65°.
Now, in △ADC,
=> CD is the altitude to AB.
So,
=> ADC = 90°
=> CAD =
CAB = 65°.
So,
=> ADC +
CAD +
DCA = 180° (Angle sum Property.)
=> 90° + 65° + DCA = 180°
=> 155° +DCA = 180°
=> DCA = 180° - 155°
=> DCA = 25°
Hence, we get,
Answer:
+
*LN(|
|) +C
Step-by-step explanation:
we will have to do a trig sub for this
use x=a*tanθ for sqrt(x^2 +a^2) where a=2
x=2tanθ, dx= 2 sec^2 (θ) dθ
this turns into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ
the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1
then it simplifies into integral(4*sec^3 (θ)) dθ
you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C
then you will need to rework your functions of θ back into functions of x
tanθ will resolve back into (see substitutions) while secθ will resolve into
sec(θ)= is from its ratio identity of hyp/adj where the hyp. is
and adj is 2 (see tan(θ) ratio)
after resolving back into functions of x, substitute ratios for trig functions:
= +
*LN(|
|) +C