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The School of Athens
Euclid, as imagined by Raphael in this detail from The School of Athens.
Jamshed Al-Kashi
Al-Kashi was one of the greatest mathematicians of the 15th century, calculating the sin(1*) correct to 16 decimal places.
Al-Beruni
The 11th century scientist and philosopher managed to estimate the radius of the earth using a very creative method that employed trigonometry.
15 comments:
Could someone give me a hint or two for number 8? I am not sure how to go from angle to another with only the sin.
Also, for number 9 do you really need length b? I was thinking that you could find the angle, then the sign and therefore the tan. Maybe I am wrong...
scratch my last post about 8. I figured it out.
It has completely slipped my mind but how would you go about #1? Can somebody just give me a push in the direction, I seem to have totally forgotten.
Steven. I used the two sides given to find c. Then I used the ratios we have been using to find sin(v). (I used a and c in my ratios.) hope this helps.
I'm having trouble with 7. Is sin(v) side a? And if so, how would you find the answer for 7?
Could someone please help me with #4?
Oops not #4, #6 please.
Never mind! I figured those out, but if someone could help on 10 and 11 that would be great.
Lauren, for numbers 10 and 11 I started by drawing an altitude for the triangle. You now have two right triangles that you can use to find the altitudes length and then the area. Does this help?
how did someone go about working on number 3?
Chap, I used the sin-1(2/4) to find angle u. Once you find the sine of angle u, you can find the angle. It is a sine that for which we have the angle. Hope this helps.
chap- i used the ratio for hypotenuse over opposite side and then used the inverse sin function
i am still a little confused on 10 and 11- victoria how did you find the sides of the two triangles you made?
Hope this is not too late. For number 10, I used the sin(25)/1=x/15. x is the altitude. Now you have the height, and you can find the area.
Number 11. I keept the altitude, but not I have two 30, 60, 90 triangles. I know that the altitude is half of the hypotenuse. From there you can find the two sides making up the base, add them together, divide by two, and multiply by the hypotenuse to find the area. Does this help?
I thoght that for number eleven there would be one 30,60,90 triangle and one 45,45,90 triangle...
Lauren, you are right. My bad.
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