Question

Can anybody solve this geometric problem by using trigonometry?
solve for the angle ? please don't scam.​

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

refer the named diagram uploaded above

I don't know is my answer ryt or wrong

but have tried my level hard

I would suggest you that from next time onwards

pls try to give entire question

it's very confusing without entire question

Step-by-step explanation:

so here let the given quadrilateral be named as ABCD

so here quadrilateral ABCD is a square

so thus AB=BC=DC=AD (1)

so now let the smaller triangle named as MBC

so here angle MBC=angle MCB=15 degrees

so thus remaining angle ie angle BMC=150 degrees

so here angle BMC is the biggest angle

so we know that side opp. to biggest angle is biggest side

hence BC is our biggest side

thus then BC is our hypotenuse

so now sin MBC=MC/BC

now sin 15 can be written as sin(45-30)

so here sin(45-30) is in form sin(A-B)

so we know that sin(A-B)=sinA.cos B-sin B.cos A

here A=45 B=30

thus

sin 15=sin 45.cos 30-sin 30.cos 45

so we know that

sin 45=1/√2

cos 30=√3/2

cos 45=1/√2

sin 30=1/2

thus then

=(1/√2×√3/2)-(1/2×1/√2)

=√3/2√2-1/2√2

=√3-1//2√2

so as we know that

√3=1.73

√2=1.41

hence

=1.73-1/2×1.41

=0.73/2.82

=0.2588

ie approx 0.25

so sin 15=0.25

so then

0.25=MC/BC

ie MC=0.25.BC (2)

so now consider triangle DCM

here looking at the figure

we can say that CD is the biggest side

thus then CD is our hypotenuse

so then

sin MDC=MC/CD

=MC/BC from (1)

=0.25 BC/BC from (2)

=0.25

So we know that

sin 15=0.25

hence angle MDC=15 degrees